Theorem If $G$ is a nilpotent group and $H < G$ then $H < N_G(H)$.
proof: Take its central series. Let $i$ be minimal such that $Z_i(G) \not \le H$ then certainly $i > 0$ and $Z_{i-1}(G) \le H$. We know $Z_{i-1} \lhd Z_i$, take $z \in Z_i(G) \setminus H$ then $Z_{i-1}(G)z$ lies in the center of $G/Z_{i-1}(G)$ (since $Z_i/Z_{i-1} \le Z(G/Z_{i-1}(G))$). For all $h \in H \le G$ we get $zh Z_{i-1}(G) = hz Z_{i-1}(G)$ and since $Z_{i-1}(G) \le H$ we have $[h,z] \in H$ for all $h$: by basic properties of commutators [and that $z$ is not in $H$] this shows that $z \in N_G(H) \setminus H$.
Proposition (This is a Frattini argument) If $H \unlhd G$ and $P$ is a Sylow $p$-subgroupof $H$ for some prime $p$ then $G = N_G(P) H$.
proof: Take $g \in G$ and consider that $P^g \le H^g = H$ is a Sylow $p$-group hence conjugate to $P$, i.e. there's some $h \in H$ such that $P^g = P^h$ which implies $gh^{-1} \in N_G(P)$ so each $g \in N_G(P) H$.
Theorem If $P$ is a Sylow $p$-subgroup of $G$ then $N_G(P) = N_G(N_G(P))$.
proof: $P$ is the only Sylow $p$-subgroup of $N_G(P)$! Any other would be conjugate (by Sylow's theorem) but $P^g = P$ by the definition of normalizer. $P \le N_G(P) \le N_G(N_G(P))$ so in fact $P$ is the only Sylow $p$-group of $N_G(N_G(P))$ too, we just need to show that $N_G(N_G(P)) \le N_G(P)$ now: let $x \in N_G(N_G(p))$ so $N_G(P)^x = N_G(P)$ and hence $P^x \le N_G(P)$ but we've already seen that conjugates of Sylow $p$-groups are all equal so $x \in N_G(P)$.
proof: You can prove it quicker with Frattini: $N_G(P) \unlhd N_G(N_G(P))$ so $N_G(N_G(P)) = N_G(P)N_{N_G(P)}(P)$ but $N_{N_G(P)}(P) \le N_G(P)$ so $N_G(N_G(P)) = N_G(P)$.
Theorem If $H$ is a subgroup of $G$ which contains the normalizer of a Sylow subgroup $P$, then $H=N_G(H)$.
proof: Since $H \unlhd N_G(H)$ the Frattini argument gives us $N_G(H) = N_{N_G(H)}(P) H$ but $N_{N_G(H)}(P) \le N_{G}(P) \le H$.
Proposition For $p$ prime, a nilpotent group is a direct product of its $p$-groups.
proof: We have already seen that a $p$-group is nilpotent and the direct product of nilpotent groups is nilpotent. In the other direction, let $G$ be some nilpotent group and take any $p$-Sylow subgroup of $G$ for some $p$ dividing $|G|$: let $N=N_G(P)$ we know $N < G$ implies $N < N_G(N)$ but this contradicts the previous result, so $N=G$ thus $P \unlhd G$, since all the $p$-groups have coprime order their direct product gives back the original group. (Note: $G = N_G(P)$ implies $P \unlhd G$ is proved in the properties about normalizers).
Theorem Normalizers always group in a nilpotent group.
proof:
Let $H$ be a subgroup of a nilpotent group $G$, then if $H$ doesn't
contain the center take some element of the center not in $H$: that'll
be in the normalizer. If $H$ is the center the normalizer will be the
whole group. If $H$ strictly contains the (nontrivial!) center $Z$ then by induction (which we can only do due to the nilpotent condition) the normalizer of $H/Z$ in $G/Z$ will grow and there will be some $nZ$ in the normalizer which isn't of the form $hZ$, and $(H/Z)^(nZ) = H/Z$ so $H^n=H$.
Tuesday 29 January 2013
Saturday 26 January 2013
Optimal series for nilpotent groups
Definition $\Gamma_1(G) = G$ and $\Gamma_{i+1}(G) = [\Gamma_i(G),G]$.
Definition $Z_0(G) = 1$ and $Z_i(G)$ is defined to be the unique (prove this) group sch that $Z_i(G)/Z_{i-1}(G) = Z(G/Z_{i-1}(G))$.
Note: This is the biggest possible group that $Z_i(G)$ could be and it still satisfying the condition needed for a central series.
Definition lower central series $$G = \Gamma_1(G) \unrhd \Gamma_2(G) \unrhd \Gamma_3(G) \unrhd \cdots.$$
Definition upper central series $$1 = Z_0(G) \unlhd Z_1(G) \unlhd Z_2(G) \unlhd \cdots.$$
Proposition $\Gamma_2(G) = G' = G^{(1)}$ and $Z_1(G) = Z(G)$ and all $Z$-terms are actually characteristic subgroups.
Theorem $G$ is nilpotent iff $\Gamma_n(G) = 1$ (for some n) iff $Z_n(G) = G$ (for some n), furthermore if $G$ had a central series $$1 = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_r = G$$ then we have (the optimality conditions) for each $0 \le i \le r$, $\Gamma_{r-i+1}(G) \le G_{i} \le Z_i(G)$ and finally $\Gamma_{c+1}(G) = 1$ iff $Z_c(G) = G$.
proof: Certainly if either $\Gamma_{n+1}(G) = 1$ iff $Z_n(G) = G$ for some $n$ then the lower or upper series are a central series and $G$ is nilpotent. Now suppose $G$ is nilpotent and take a central series we'll prove the optimality conditions separately by induction.
First $\Gamma_{i+1}(G) \le G_{r-i}$, assume it's true for $i$. then $[\Gamma_{i+1}(G),G] \le [G_{r-i},G] \le G_{r-i-1}$ done.
Second $G_{i} \le Z_i(G)$, to get $G_{i+1} \le Z_{i+1}(G) = Z(G/Z_i(G))$ we'll show that $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$ by showing for arbitrary elements $l \in G_{i+1}$ and $g \in G$ that $lZ_i(g)$ commutes with $gZ_i(G)$... but $[G_{i+1},G] \le G_i \le Z_i(G)$ so we are done! Explaining this a bit more: for any $x \in G_{i+1}$ and $y \in G$ we have $[x,y] \in Z_i(G)$, that's the same as $x Z_i(G)$ commuting with $y Z_i(G)$, which is the same as $xZ_i(G)$ being in the center $Z(G/Z_i(G))$ i.e. $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$.
Finally suppose $\Gamma_{c+1}(G)=1$, by taking $G_j = \Gamma_{c-j+1}(G)$ for each $0 \le j \le c$ and $r=c$ we have $G = \Gamma_1(G) = G_c \le Z_c(G)$ so $g = Z_c(G)$. On the other hand if $Z_c(G) = G$ then by taking $G_j = Z_j(G)$ we have $\Gamma_{c+1}(G) \le G_0 = 1$ so $\Gamma_{c+1}(G) = 1$.
Definition The smallest $n$ such that $Z_n(G) = G$ is the nilpotency class.
todo: prove it's the same for gamma and and other series.
Definition $Z_0(G) = 1$ and $Z_i(G)$ is defined to be the unique (prove this) group sch that $Z_i(G)/Z_{i-1}(G) = Z(G/Z_{i-1}(G))$.
Note: This is the biggest possible group that $Z_i(G)$ could be and it still satisfying the condition needed for a central series.
Definition lower central series $$G = \Gamma_1(G) \unrhd \Gamma_2(G) \unrhd \Gamma_3(G) \unrhd \cdots.$$
Definition upper central series $$1 = Z_0(G) \unlhd Z_1(G) \unlhd Z_2(G) \unlhd \cdots.$$
Proposition $\Gamma_2(G) = G' = G^{(1)}$ and $Z_1(G) = Z(G)$ and all $Z$-terms are actually characteristic subgroups.
Theorem $G$ is nilpotent iff $\Gamma_n(G) = 1$ (for some n) iff $Z_n(G) = G$ (for some n), furthermore if $G$ had a central series $$1 = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_r = G$$ then we have (the optimality conditions) for each $0 \le i \le r$, $\Gamma_{r-i+1}(G) \le G_{i} \le Z_i(G)$ and finally $\Gamma_{c+1}(G) = 1$ iff $Z_c(G) = G$.
proof: Certainly if either $\Gamma_{n+1}(G) = 1$ iff $Z_n(G) = G$ for some $n$ then the lower or upper series are a central series and $G$ is nilpotent. Now suppose $G$ is nilpotent and take a central series we'll prove the optimality conditions separately by induction.
First $\Gamma_{i+1}(G) \le G_{r-i}$, assume it's true for $i$. then $[\Gamma_{i+1}(G),G] \le [G_{r-i},G] \le G_{r-i-1}$ done.
Second $G_{i} \le Z_i(G)$, to get $G_{i+1} \le Z_{i+1}(G) = Z(G/Z_i(G))$ we'll show that $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$ by showing for arbitrary elements $l \in G_{i+1}$ and $g \in G$ that $lZ_i(g)$ commutes with $gZ_i(G)$... but $[G_{i+1},G] \le G_i \le Z_i(G)$ so we are done! Explaining this a bit more: for any $x \in G_{i+1}$ and $y \in G$ we have $[x,y] \in Z_i(G)$, that's the same as $x Z_i(G)$ commuting with $y Z_i(G)$, which is the same as $xZ_i(G)$ being in the center $Z(G/Z_i(G))$ i.e. $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$.
Finally suppose $\Gamma_{c+1}(G)=1$, by taking $G_j = \Gamma_{c-j+1}(G)$ for each $0 \le j \le c$ and $r=c$ we have $G = \Gamma_1(G) = G_c \le Z_c(G)$ so $g = Z_c(G)$. On the other hand if $Z_c(G) = G$ then by taking $G_j = Z_j(G)$ we have $\Gamma_{c+1}(G) \le G_0 = 1$ so $\Gamma_{c+1}(G) = 1$.
Definition The smallest $n$ such that $Z_n(G) = G$ is the nilpotency class.
todo: prove it's the same for gamma and and other series.
Solvable simple groups have prime order!
Theorem The solvable simple groups have prime order.
proof: Let $G$ be solvable then $G' = [G,G] \lhd G$ by the optimal-series theory, if $G$ is simple then $G' = 1$ so $G$ is abelian and the result follows from...
Lemma The abelian simple groups are the cyclic groups of prime order.
proof: By Lagrange's theorem if a group has prime order it has no nontrivial subgroups. Every subgroup of an abelian group is normal, so we need to classify the abelian groups that have no nontrivial subgroups. The group must be cyclic since we need every element to be a generator (except 1) so we see it's at least $C_{p^r}$ but $C_{p^{r-1}}$ (by raising everything to the $p$th power) is a subgroup of that and that's only nontrivial when $r=1$.
Theorem $G$ is solvable iff every composition factor has prime power order.
proof: If $G$ is solvable then any factor of its series is solvable by basic-properties, but such factors are also simple because we have a composition series (one which admits no refinement), so we are done by the above theorem.
Definition A subgroup $N$ of $G$ is a minimal normal subgroup if $N \unlhd G$ and for any $M \unlhd G$ with $M \le N$ either $1 = M$ or $M = N$.
Definition A group is an elementary abelian $p$-group if it is a direct product of copies of the cyclic group $C_p$.
Proposition A minimal normal $N$ subgroup of a solvable group is elementary abelian.
proof: Since it's a subgroup of a solvable group $N$ is solvable too, so it's derived series terminates thus $N' = [N,N] < N$, but $N' \text{ char } N$ and $N \lhd G$ so $N' \lhd G$ implies $N' = 1$. Therefore $N$ is abelian so by the structure theorem it's a product of cylic groups, let $p \mid |N|$ and consider the normal subgroup $M$ of order $p$ elements: $M\text{ char }N$ so $M=N$, every factor is $C_p$.
proof: Let $G$ be solvable then $G' = [G,G] \lhd G$ by the optimal-series theory, if $G$ is simple then $G' = 1$ so $G$ is abelian and the result follows from...
Lemma The abelian simple groups are the cyclic groups of prime order.
proof: By Lagrange's theorem if a group has prime order it has no nontrivial subgroups. Every subgroup of an abelian group is normal, so we need to classify the abelian groups that have no nontrivial subgroups. The group must be cyclic since we need every element to be a generator (except 1) so we see it's at least $C_{p^r}$ but $C_{p^{r-1}}$ (by raising everything to the $p$th power) is a subgroup of that and that's only nontrivial when $r=1$.
Theorem $G$ is solvable iff every composition factor has prime power order.
proof: If $G$ is solvable then any factor of its series is solvable by basic-properties, but such factors are also simple because we have a composition series (one which admits no refinement), so we are done by the above theorem.
Definition A subgroup $N$ of $G$ is a minimal normal subgroup if $N \unlhd G$ and for any $M \unlhd G$ with $M \le N$ either $1 = M$ or $M = N$.
Definition A group is an elementary abelian $p$-group if it is a direct product of copies of the cyclic group $C_p$.
Proposition A minimal normal $N$ subgroup of a solvable group is elementary abelian.
proof: Since it's a subgroup of a solvable group $N$ is solvable too, so it's derived series terminates thus $N' = [N,N] < N$, but $N' \text{ char } N$ and $N \lhd G$ so $N' \lhd G$ implies $N' = 1$. Therefore $N$ is abelian so by the structure theorem it's a product of cylic groups, let $p \mid |N|$ and consider the normal subgroup $M$ of order $p$ elements: $M\text{ char }N$ so $M=N$, every factor is $C_p$.
Optimal series for solvable groups
In this section we will use potentially infinite series which may or may not terminate, this forces us to distinguish between ascending and descending series.
Definition Set $G^{(0)} := G$ and for each $i$ define $G^{(i)} = (G^{(i)})' = [G^{i-1},G^{i-1}]$. This gives us the derived (descending) series $$G = G^{(0)} \unrhd G^{(1)} \unrhd G^{(2)} \unrhd G^{(3)} \unrhd \cdots.$$
Prop $G^{(i)} \operatorname{char} G$. proof: unproved
Theorem $G$ is solvable iff the derived series terminates with $G^{(n)}=1$, furthermore it is optimal in the following sense if $$(\star)\,, 1 = G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_r = G$$ then we have $G^{(i)} \le G_{r-i}$.
proof: By the commutator characterization of abelian series, each factor of this derived series is abelian. So if $G^{(n)} = 1$ for some $n$ then we have an abelian series and $G$ is solvable. For the converse, suppose $G$ is solvable, let $(\star)$ be an abelian series for it. We will show by induction that $G^{(i)} \le G_{r-i}$. Suppose it is true for $i$, then $G^{(i+1)} \le [G^{(i)},G^{(i)}] \le [G_{r-i},G_{r-i}] = (G_{r-i})' \le G_{r-i-1}$ again by the commutator characterization.
Corollary Every solvable group has an abelian normal series.
Definition If $G$ is solvable the minimal length of series is called it's derived length. show this is the same as the length of the derived series?
Definition Set $G^{(0)} := G$ and for each $i$ define $G^{(i)} = (G^{(i)})' = [G^{i-1},G^{i-1}]$. This gives us the derived (descending) series $$G = G^{(0)} \unrhd G^{(1)} \unrhd G^{(2)} \unrhd G^{(3)} \unrhd \cdots.$$
Prop $G^{(i)} \operatorname{char} G$. proof: unproved
Theorem $G$ is solvable iff the derived series terminates with $G^{(n)}=1$, furthermore it is optimal in the following sense if $$(\star)\,, 1 = G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_r = G$$ then we have $G^{(i)} \le G_{r-i}$.
proof: By the commutator characterization of abelian series, each factor of this derived series is abelian. So if $G^{(n)} = 1$ for some $n$ then we have an abelian series and $G$ is solvable. For the converse, suppose $G$ is solvable, let $(\star)$ be an abelian series for it. We will show by induction that $G^{(i)} \le G_{r-i}$. Suppose it is true for $i$, then $G^{(i+1)} \le [G^{(i)},G^{(i)}] \le [G_{r-i},G_{r-i}] = (G_{r-i})' \le G_{r-i-1}$ again by the commutator characterization.
Corollary Every solvable group has an abelian normal series.
Definition If $G$ is solvable the minimal length of series is called it's derived length. show this is the same as the length of the derived series?
Thursday 24 January 2013
Basic properties of nilpotent and solvable groups (unfinished)
Theorem If $G$ is nilpotent or solvable so is any subgroup.
proof: Let $G$ have the series $$1 = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G.$$ Suppose $H \le G$ then $$1 = G_0 \cap H \lhd G_1 \cap H \lhd \cdots \lhd G_n \cap H = H$$ is a series for $H$ by ABC.
Now suppose $G$ was nilpotent, by the alternative characterization of a central factor we have that $[G,G_i] \le G_{i-1}$ and we deduce $H$ is nilpotent from $[G \cap H,G_i \cap H] \le G_{i-1} \cap H$.
Now suppose $G$ was solvable we deduce that $H$ is too from the fact that if $G_{i}/G_{i-1}$ is abelian so is $G_{i} \cap H/G_{i-1} \cap H \simeq G_{i-1}(G_i \cap H)/G_{i-1} \le G_i/G_{i-1}$ (by ABC) as it's the subgroup of an abelian group.
Theorem If $G$ is nilpotent or solvable so is any quotient.
proof: Suppose $G$ was solvable, $N \unlhd G$ and let $\nu : G \to G/N$ be the natural inclusion homomorphism which we know to be surjective. Take $G$s composition series as before. Let us call $\nu(G_i)$ by $\bar H_i$ so by Noether(-1) and the fact it's a composition series $\bar H_i \lhd \bar H_{i+1}$ (so we have a series for $G/N$) also it's an abelian series since $\bar H_{i+1}/\bar H_i$ is abelian: let $\bar x, \bar y \in \bar H_{i+1}$ say $\bar x = \nu x$, $\bar y = \nu y$ then since $H_{i+1}/ H_i$ is abelian $xy = yxd$ for some $d \in H_i$, so taking $\nu$ we find $\bar x \bar y = \bar y \bar x \bar d$ for some $\bar d \in \bar H_i$, thus: $$(\bar x \bar H_i)(\bar y \bar H_i) = \bar y \bar x \bar d \bar H_i = \bar y \bar x \bar H_i = \bar y \bar H_i \bar x \bar H_i.$$
Theorem If $H$ and $K$ are both nilpotent or both solvable so is $H\times K$
proof:
Proposition Nilpotent groups aren't closed under taking extensions. Example $S_3$ is an extension of $A_3 \simeq C_3$ by $S_3/A_3 \simeq C_2$ but isn't nilpotent.
Theorem Solvable groups are closed under taking extensions: Suppose $K \unlhd G$ then if $K$ and $G/K$ are both solvable so is $G$.
proof:
proof: Let $G$ have the series $$1 = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G.$$ Suppose $H \le G$ then $$1 = G_0 \cap H \lhd G_1 \cap H \lhd \cdots \lhd G_n \cap H = H$$ is a series for $H$ by ABC.
Now suppose $G$ was nilpotent, by the alternative characterization of a central factor we have that $[G,G_i] \le G_{i-1}$ and we deduce $H$ is nilpotent from $[G \cap H,G_i \cap H] \le G_{i-1} \cap H$.
Now suppose $G$ was solvable we deduce that $H$ is too from the fact that if $G_{i}/G_{i-1}$ is abelian so is $G_{i} \cap H/G_{i-1} \cap H \simeq G_{i-1}(G_i \cap H)/G_{i-1} \le G_i/G_{i-1}$ (by ABC) as it's the subgroup of an abelian group.
Theorem If $G$ is nilpotent or solvable so is any quotient.
proof: Suppose $G$ was solvable, $N \unlhd G$ and let $\nu : G \to G/N$ be the natural inclusion homomorphism which we know to be surjective. Take $G$s composition series as before. Let us call $\nu(G_i)$ by $\bar H_i$ so by Noether(-1) and the fact it's a composition series $\bar H_i \lhd \bar H_{i+1}$ (so we have a series for $G/N$) also it's an abelian series since $\bar H_{i+1}/\bar H_i$ is abelian: let $\bar x, \bar y \in \bar H_{i+1}$ say $\bar x = \nu x$, $\bar y = \nu y$ then since $H_{i+1}/ H_i$ is abelian $xy = yxd$ for some $d \in H_i$, so taking $\nu$ we find $\bar x \bar y = \bar y \bar x \bar d$ for some $\bar d \in \bar H_i$, thus: $$(\bar x \bar H_i)(\bar y \bar H_i) = \bar y \bar x \bar d \bar H_i = \bar y \bar x \bar H_i = \bar y \bar H_i \bar x \bar H_i.$$
Theorem If $H$ and $K$ are both nilpotent or both solvable so is $H\times K$
proof:
Proposition Nilpotent groups aren't closed under taking extensions. Example $S_3$ is an extension of $A_3 \simeq C_3$ by $S_3/A_3 \simeq C_2$ but isn't nilpotent.
Theorem Solvable groups are closed under taking extensions: Suppose $K \unlhd G$ then if $K$ and $G/K$ are both solvable so is $G$.
proof:
- crazyproject: subgroups-and-quotient-groups-of-solvable-groups-are-solvable
- groupprops/Solvable_group#Metaproperties
- Schaum's Outline of Group Theory - B. Baumslag (Author), B. Chandler (Author)
Tuesday 22 January 2013
Examples 1
- $1 \lhd K \lhd A_4 \lhd S_4$ is not a normal series because $K$ is not normal in $S_4$?? $K/1 = C_2$? $V/K = C_2$, $A_4/V = C_3$, $S_4/A_4 = C_2$
- The group $S_5$ is not solvable [washington link]
- The commutator subgroup of the alternating group A4 is the Klein four group.
- The commutator subgroup of the symmetric group Sn is the alternating group An.
- The commutator subgroup of the quaternion group Q = {1, −1, i, −i, j, −j, k, −k} is [Q,Q]={1, −1}.
- A=<a>,B=<b> subgroups of D6 but AB is not a group, because ab is in it but the inverse (ab)^-1 = abab is not.
- Theorem $t = (1\,2)$ and $c = (1\,2\,3\,\ldots\,n)$ generate $S_n$.
proof: $t^{c^i} = (i\,i+1)$ so we have every transposition. $t^{c^{n-1}}c = (n-1\,n)c = (1\,2\,3\,\ldots\,n-1)$ so we are done by induction.
Commutator Subgroups
Definition The commutator $[g,h] = g^{-1} h^{-1} g h$ has the following properties $[h,g] = [g,h]^{-1}$, $[g,h] = 1$ iff $gh = hg$.
Proposition If $[x,y] \in G$ then $xG$ and $yG$ commute.
proof: $xGyG = xyG$ by the definition of coset groups. $yxG = yx(x^{-1}y^{-1}xy)G = xyG$ because $G = x^{-1}y^{-1}xyG$ since $x^{-1}y^{-1}xy \in G$.
Lemma If $[g,z] \in G$ then $z^{-1} g z G = g G = G$ then $g^z \in G$.
Definition If $H,K \le G$ their commutator $[H,K] = [K,H] = \langle [h,k] \in G|h \in H, g \in G \rangle$ (Note! This is the group generated by commutators, not just the set of all commutators).
Definition The commutator subgroup or derived group of $G$ is $[G,G]$.
Proposition $[G,G] \text{char} G$
proof: Given $\alpha \in Aut(G)$, $[g,h]^\alpha = [g^\alpha,h^\alpha]$.
Proposition (The commutator subgroup is the smallest normal subgroup $N$ such that $G/N$ is abelian) Given $N \unlhd G$ then $G/N$ is abelian iff $[G,G] \le N$.
proof: (for all $g$,$h$ in $G$) $Ng Nh = Nh Ng$ iff $Ngh = Nhg$ iff $[g^{-1},h^{-1}] \in N$.
Theorem if $A \le B$ are both subgroups of $G$ then $[A,G] \le [B,G]$.
proof: We show the stronger result that the generators of $[A,G]$ are contained in the generators of $[B,G]$ and this is just immediate from $A$ being a subset of $B$.
Theorem A factor $G_i/G_{i-1}$ of a series for $G$ is central iff $[G,G_i] \le G_{i-1}$
proof: todo
Theorem Any index 2 subgroup is normal.
proof: Let $H$ have index 2 in $G$, then there are exactly 2 cosets of it. For $a \in H$ we $aH=H=Ha$ so the other coset must be all the other elements so for $b \in H$, $bH=Hb$.
Theorem Any index 2 subgroup contains the commutator subgroup.
proof: If $H$ has index 2 then $G/H \simeq C_2$ is abelian so $H$ contains $G'$.
Corollary The only index 2 subgroup of $S_n$ is $A_n$.
Proposition If $[x,y] \in G$ then $xG$ and $yG$ commute.
proof: $xGyG = xyG$ by the definition of coset groups. $yxG = yx(x^{-1}y^{-1}xy)G = xyG$ because $G = x^{-1}y^{-1}xyG$ since $x^{-1}y^{-1}xy \in G$.
Lemma If $[g,z] \in G$ then $z^{-1} g z G = g G = G$ then $g^z \in G$.
Definition If $H,K \le G$ their commutator $[H,K] = [K,H] = \langle [h,k] \in G|h \in H, g \in G \rangle$ (Note! This is the group generated by commutators, not just the set of all commutators).
Definition The commutator subgroup or derived group of $G$ is $[G,G]$.
Proposition $[G,G] \text{char} G$
proof: Given $\alpha \in Aut(G)$, $[g,h]^\alpha = [g^\alpha,h^\alpha]$.
Proposition (The commutator subgroup is the smallest normal subgroup $N$ such that $G/N$ is abelian) Given $N \unlhd G$ then $G/N$ is abelian iff $[G,G] \le N$.
proof: (for all $g$,$h$ in $G$) $Ng Nh = Nh Ng$ iff $Ngh = Nhg$ iff $[g^{-1},h^{-1}] \in N$.
Theorem if $A \le B$ are both subgroups of $G$ then $[A,G] \le [B,G]$.
proof: We show the stronger result that the generators of $[A,G]$ are contained in the generators of $[B,G]$ and this is just immediate from $A$ being a subset of $B$.
Theorem A factor $G_i/G_{i-1}$ of a series for $G$ is central iff $[G,G_i] \le G_{i-1}$
proof: todo
Theorem Any index 2 subgroup is normal.
proof: Let $H$ have index 2 in $G$, then there are exactly 2 cosets of it. For $a \in H$ we $aH=H=Ha$ so the other coset must be all the other elements so for $b \in H$, $bH=Hb$.
Theorem Any index 2 subgroup contains the commutator subgroup.
proof: If $H$ has index 2 then $G/H \simeq C_2$ is abelian so $H$ contains $G'$.
Corollary The only index 2 subgroup of $S_n$ is $A_n$.
The ABC theorem
Theorem Let $A,B,C$ be subgroups of $G$. Let $B \le A \le G$ and $C \le G$, then we have:
A) $A \cap BC = B(A \cup C)$ as sets.
B) if $B \unlhd A$ then $B \cap C \unlhd A \cap C$ and $(A \cap C)/(B\cap C) \simeq B(A\cap C)/B$
C) If $B \unlhd A$ and $C \unlhd G$ then $BC \unlhd AC$ and $AC/BC \simeq A/B(A\cap C)$.
proof hint: Use Noethers isomorphism theorems and sometimes consider elements.
A) $A \cap BC = B(A \cup C)$ as sets.
B) if $B \unlhd A$ then $B \cap C \unlhd A \cap C$ and $(A \cap C)/(B\cap C) \simeq B(A\cap C)/B$
C) If $B \unlhd A$ and $C \unlhd G$ then $BC \unlhd AC$ and $AC/BC \simeq A/B(A\cap C)$.
proof hint: Use Noethers isomorphism theorems and sometimes consider elements.
Central Series and the Nilpotent p-group
Definition $H/K$ is a central factor of a series $\ldots \unlhd K \unlhd H \ldots \unlhd G$ if $K \unlhd G$ and $H/K \le Z(G/K)$.
Lemma In a series such as the above, H/K is central iff $[H,G] \le K$.
proof: ($\Rightarrow$) Suppose $H/K$ is a central factor so $K \unlhd G$ and $H/K \le Z(G/K)$ then for all $g \in G$, $h \in H$, $KgKh = KhKg$ so $ghK=hgK$ and $h^{-1}g^{-1}hgK=K$ whence $[h,g] \in K$ thus $[H,G] \le K$. ($\Leftarrow$) very similar.. unproved
Definition A series with all factors central is a central series.
Definition A group with a central series is nilpotent.
Definition A series with all factors abelian is an abelian series.
Definition A group with an abelian series is solvable. (This may be familiar from Galois theory)
Definition A $p$-group is one whose order is a power of the prime $p$.
Lemma If $G$ is a nontrivial $p$-group then $Z(G)$ is also non-trivial.
proof: Each conjugacy class has size some power of $p$ and by the Conjugacy Class Formula the sum of these is $|G|$. The identity lies in a class of size $1$ so there must be at least $p-1$ other such conjugacy classes in $Z(G)$.
Theorem A group of order $p^2$ is abelian.
proof: A nontrivial element of $Z(G)$ either generates the whole group $C_{p^2}$ or generates $C_p \unlhd G$ in which case some element of $G$ not in that $C_p$ will generate the rest of group and $G$ is $C_p \times C_p$.
This same argument doesn't work for $p^3$ or higher: note $D_8$ is non-abelian, it's center is $C_2$ not the whole group. If $G$ is abelian then $Z(G)=G$.
Theorem A $p$-group is nilpotent.
proof: Let $G$ be a $p$-group and we will proceed by induction on $|G|$. Let $Z = Z(G)$. So $G/Z(G)$ has a central series which by Noether3,4 is, up to isomorphism, of the form $$Z/Z = G_0/Z \unlhd G_1/Z \unlhd \cdots \unlhd G_n/Z = G/Z$$ with $(G_i/Z)/(G_{i-1}/Z)$ central (i.e. a subgroup of $Z((G/Z)/(G_{i-1}/Z))$) therefore (by Noether4) $G_i/G_{i-1} \le Z(G/G_i)$ and $$Z = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_n = G$$ is a central series for our $p$-group if you stick 1 on the left.
Lemma In a series such as the above, H/K is central iff $[H,G] \le K$.
proof: ($\Rightarrow$) Suppose $H/K$ is a central factor so $K \unlhd G$ and $H/K \le Z(G/K)$ then for all $g \in G$, $h \in H$, $KgKh = KhKg$ so $ghK=hgK$ and $h^{-1}g^{-1}hgK=K$ whence $[h,g] \in K$ thus $[H,G] \le K$. ($\Leftarrow$) very similar.. unproved
Definition A series with all factors central is a central series.
Definition A group with a central series is nilpotent.
Definition A series with all factors abelian is an abelian series.
Definition A group with an abelian series is solvable. (This may be familiar from Galois theory)
Definition A $p$-group is one whose order is a power of the prime $p$.
Lemma If $G$ is a nontrivial $p$-group then $Z(G)$ is also non-trivial.
proof: Each conjugacy class has size some power of $p$ and by the Conjugacy Class Formula the sum of these is $|G|$. The identity lies in a class of size $1$ so there must be at least $p-1$ other such conjugacy classes in $Z(G)$.
Theorem A group of order $p^2$ is abelian.
proof: A nontrivial element of $Z(G)$ either generates the whole group $C_{p^2}$ or generates $C_p \unlhd G$ in which case some element of $G$ not in that $C_p$ will generate the rest of group and $G$ is $C_p \times C_p$.
This same argument doesn't work for $p^3$ or higher: note $D_8$ is non-abelian, it's center is $C_2$ not the whole group. If $G$ is abelian then $Z(G)=G$.
Theorem A $p$-group is nilpotent.
proof: Let $G$ be a $p$-group and we will proceed by induction on $|G|$. Let $Z = Z(G)$. So $G/Z(G)$ has a central series which by Noether3,4 is, up to isomorphism, of the form $$Z/Z = G_0/Z \unlhd G_1/Z \unlhd \cdots \unlhd G_n/Z = G/Z$$ with $(G_i/Z)/(G_{i-1}/Z)$ central (i.e. a subgroup of $Z((G/Z)/(G_{i-1}/Z))$) therefore (by Noether4) $G_i/G_{i-1} \le Z(G/G_i)$ and $$Z = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_n = G$$ is a central series for our $p$-group if you stick 1 on the left.
Normalizers and the Conjugacy-Class equation
Definition The normalizer $N_G(S)$ of a subset $S$ of a group $G$ is the set of all $a \in G$ such that $S^a = S$.
Theorem $N_G(S) \le G$.
proof: Clearly $1 \in N_G(S)$. Let $a,b$ in $N_G(S)$ then $S^{ab} = (S^a)^b = S^b = S$ so $ab$ is too. Since the group is finite we have inverses too (make n big enough so that $a^n = 1$ then $a^{-1} = a^{n-1}$). Alternative (more sensible) argument: $S^{1} = S^{aa^{-1}}=S^{a^{-1}}$.
Theorem If $H \le G$ then $H \unlhd N_G(H)$.
proof: We have already seen that it's a group, we just need to show $\forall n \in N_G(H), nH = Hn$ but that's immediate by the definition of normalizer.
Corollary When $H \le G$, $G = N_G(H)$ implies $H \unlhd G$.
Definition Another equivalence relation, conjugacy is defined by $x \sim y$ if there exists some $g$ such that $x^{g} = y$.
Lemma Central elements partition the group into trivial conjugacy classes.
proof: Let $a \in Z(G)$ then $g^a = a^{-1}ga = g$ for all $g$ so every such conjugacy class is a single element.
Lemma The number of conjugates of $x$ (including $x$) is $[G:N_G(x)]$.
proof: orbit stabilizer.
Theorem (Conjugacy Class Equation) $$
\left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{x} \left[{G : N_G \left({x}\right)}\right]$$ where the sum is taken over representatives of the non-singleton conjugacy classes.
proof: This is really just a corollary of the lemma plus the information that center elements have trivial conjugacy classes, but we discuss it a bit more anyway. In the abelian case $G = Z(G)$ and there are non non-singleton conjugacy classes so the result is immediate. In the general case $G \setminus Z(G)$ are the elements which the conjugacy relation partitions into non-singleton sets - the relation proved in the previous lemma gives the equality.
Theorem $N_G(S) \le G$.
proof: Clearly $1 \in N_G(S)$. Let $a,b$ in $N_G(S)$ then $S^{ab} = (S^a)^b = S^b = S$ so $ab$ is too. Since the group is finite we have inverses too (make n big enough so that $a^n = 1$ then $a^{-1} = a^{n-1}$). Alternative (more sensible) argument: $S^{1} = S^{aa^{-1}}=S^{a^{-1}}$.
Theorem If $H \le G$ then $H \unlhd N_G(H)$.
proof: We have already seen that it's a group, we just need to show $\forall n \in N_G(H), nH = Hn$ but that's immediate by the definition of normalizer.
Corollary When $H \le G$, $G = N_G(H)$ implies $H \unlhd G$.
Definition Another equivalence relation, conjugacy is defined by $x \sim y$ if there exists some $g$ such that $x^{g} = y$.
Lemma Central elements partition the group into trivial conjugacy classes.
proof: Let $a \in Z(G)$ then $g^a = a^{-1}ga = g$ for all $g$ so every such conjugacy class is a single element.
Lemma The number of conjugates of $x$ (including $x$) is $[G:N_G(x)]$.
proof: orbit stabilizer.
Theorem (Conjugacy Class Equation) $$
\left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{x} \left[{G : N_G \left({x}\right)}\right]$$ where the sum is taken over representatives of the non-singleton conjugacy classes.
proof: This is really just a corollary of the lemma plus the information that center elements have trivial conjugacy classes, but we discuss it a bit more anyway. In the abelian case $G = Z(G)$ and there are non non-singleton conjugacy classes so the result is immediate. In the general case $G \setminus Z(G)$ are the elements which the conjugacy relation partitions into non-singleton sets - the relation proved in the previous lemma gives the equality.
Orbit-Stabilizer theorem
Definition A group action of a group on a set is just an operation $g s \in S$ for $s \in S$.
Corollary This gives an an equivalence relation on the set: $x \sim y$ if $exists g, y = gx$.
Definition The orbit ($Orb(s)$) of an element $s$ of the equivalence class containing $s$ i.e. the set is all the elements $\sim$ to $s$. Alternatively it's the canonical map $S \to S/\sim$.
Definition A group acts transitively if there is only one orbit (every element is related to every other).
Definition The stabalizer ($Stab(s))$ is the set of group elements $g$ such that $gs = s$.
Theorem $Stab(s) \le G$ (meaning it's a subgroup).
Theorem $$|Orb(x)|\cdot|Stab(s)|=|G|.$$
proof: It's also easy to see this identity holds if either of the factors are 1. First, Let $y = gx$ and $g' x = x$ then $g'^g y = g^{-1}g'gy = y$. This tells us that $|Stab(x)| = |Stab(y)|$ when $y = gx$ (because the group automorphism gives us a bijection between the elements which stabilize x and those which stabilize y), and there are |Orb(x)| such relations $x$,$y$,$\ldots$ so we have the theorem.
Corollary This gives an an equivalence relation on the set: $x \sim y$ if $exists g, y = gx$.
Definition The orbit ($Orb(s)$) of an element $s$ of the equivalence class containing $s$ i.e. the set is all the elements $\sim$ to $s$. Alternatively it's the canonical map $S \to S/\sim$.
Definition A group acts transitively if there is only one orbit (every element is related to every other).
Definition The stabalizer ($Stab(s))$ is the set of group elements $g$ such that $gs = s$.
Theorem $Stab(s) \le G$ (meaning it's a subgroup).
Theorem $$|Orb(x)|\cdot|Stab(s)|=|G|.$$
proof: It's also easy to see this identity holds if either of the factors are 1. First, Let $y = gx$ and $g' x = x$ then $g'^g y = g^{-1}g'gy = y$. This tells us that $|Stab(x)| = |Stab(y)|$ when $y = gx$ (because the group automorphism gives us a bijection between the elements which stabilize x and those which stabilize y), and there are |Orb(x)| such relations $x$,$y$,$\ldots$ so we have the theorem.
Saturday 19 January 2013
Noether4 (unfinished)
Lemma (Lattice isomorphism theorem) Let $\alpha : G \to H$ be some group homomorphism. There is a natural bijection between the (normal) subgroups of $G$ that contain $\ker(\alpha)$ and the subgroups of $\text{im}(\alpha)$.
proof: For $K \le G$ let $K^\alpha = \{ \alpha(k) \in H | k \in K \}$ (note this is a group) and for $M \le \text{im}(\alpha)$ let $M^* = \{ g \in G | \alpha(g) \in M \}$ this too is a group. Clearly $(M^*)^\alpha = \{ \alpha(g) \in G | \alpha(g) \in M \}= M$ since every element of $M$ is in the image of $\alpha$. Now $(K^\alpha)^* \le K$ easily by definition and the reverse inclusion comes down to the something about kernel. As for the inclusions unproved.
Theorem (Noether4). Let $G$ be a group and $N \unlhd G$. There is a natural bijection between the lattice of subgroups of $G$ that contain $N$ and the lattice of subgroups of $G/N$. The bijection preserves relations "subgroup", "normal subgroup" and properness.
proof: Let $H$ be a subgroup of $G$ which contains $N$ i.e. $N$ is also a subgroup of it then by Noether3 $N$ is normal in $H$ too so $H/N$ is well defined homomorphism. Now apply lattice isomorphism.
Corollary: If $B \lhd G/N$ then $\exists K, B \simeq K/N$.
proof: For $K \le G$ let $K^\alpha = \{ \alpha(k) \in H | k \in K \}$ (note this is a group) and for $M \le \text{im}(\alpha)$ let $M^* = \{ g \in G | \alpha(g) \in M \}$ this too is a group. Clearly $(M^*)^\alpha = \{ \alpha(g) \in G | \alpha(g) \in M \}= M$ since every element of $M$ is in the image of $\alpha$. Now $(K^\alpha)^* \le K$ easily by definition and the reverse inclusion comes down to the something about kernel. As for the inclusions unproved.
Theorem (Noether4). Let $G$ be a group and $N \unlhd G$. There is a natural bijection between the lattice of subgroups of $G$ that contain $N$ and the lattice of subgroups of $G/N$. The bijection preserves relations "subgroup", "normal subgroup" and properness.
proof: Let $H$ be a subgroup of $G$ which contains $N$ i.e. $N$ is also a subgroup of it then by Noether3 $N$ is normal in $H$ too so $H/N$ is well defined homomorphism. Now apply lattice isomorphism.
Corollary: If $B \lhd G/N$ then $\exists K, B \simeq K/N$.
Composition Series (unfinished)
Definition A subgroup series for a group $G$ is a chain of normal subgroups that gets down to the trivial group $$1 = G_0 \le G_1\le G_2 \le \ldots \le G_n = G.$$
Definition A series is normal if each $G_i \unlhd G_{i+1}$.
Definition The factors of a normal series are the quotients $G_{i+1}/G_{i}$.
Definition A series if proper if the containments are all strict $G_i \lhd G_{i+1}$.
Definition A series for $G$ is a refinement of another series for $G$, if it's just had zero or more new groups added in, a refinement of a normal series must be again normal. A refinement is proper if it's has at least one new group added.
Definition A composition series is a proper normal series which has no proper refinements. (so $1 \lhd G$ is a composition series for any nontrivial $G$).
Lemma Any proper normal series can be refined to a composition series.
proof: Just keep refining it until you have a composition series, a composition series must be finite since our groups are finite,.
Proposition A normal series $$1 = G_0 \unlhd G_1\unlhd G_2 \unlhd \ldots \unlhd G_n = G$$ is a composition series iff each factor is simple.
proof: ($\Leftarrow$) No factor is trivial. If a factor wasn't simple, i.e. $G_i/G_{i+1}$ has a proper nontrivial normal subgroup $B$ then $B \simeq K/G_i$ for some $G_i \lhd K \lhd G_{i+1}$ (by Noether4) and this gives a proper refinement. ($\Rightarrow$) not done yet.
Definition Two series for $G$ are equivalent if the factors of one are permutation of the factors of another.
Theorem Factors of a composition series are simple.
proof: Let us have a normal series with $\cdots \unlhd H_i \unlhd H_{i+1} \unlhd \cdots$ where $H_{i+1}/H_i$ isn't simple, say it has some normal subgroup $H_i/H_i \lhd N \lhd H_{i+1}/H_i$, then by (Noether4) lattice isomorphism we get the existence of a subgroup $H_i \lhd N' \lhd H_{i+1}$. A refinement of our series.
Definition A series is normal if each $G_i \unlhd G_{i+1}$.
Definition The factors of a normal series are the quotients $G_{i+1}/G_{i}$.
Definition A series if proper if the containments are all strict $G_i \lhd G_{i+1}$.
Definition A series for $G$ is a refinement of another series for $G$, if it's just had zero or more new groups added in, a refinement of a normal series must be again normal. A refinement is proper if it's has at least one new group added.
Definition A composition series is a proper normal series which has no proper refinements. (so $1 \lhd G$ is a composition series for any nontrivial $G$).
Lemma Any proper normal series can be refined to a composition series.
proof: Just keep refining it until you have a composition series, a composition series must be finite since our groups are finite,.
Proposition A normal series $$1 = G_0 \unlhd G_1\unlhd G_2 \unlhd \ldots \unlhd G_n = G$$ is a composition series iff each factor is simple.
proof: ($\Leftarrow$) No factor is trivial. If a factor wasn't simple, i.e. $G_i/G_{i+1}$ has a proper nontrivial normal subgroup $B$ then $B \simeq K/G_i$ for some $G_i \lhd K \lhd G_{i+1}$ (by Noether4) and this gives a proper refinement. ($\Rightarrow$) not done yet.
Definition Two series for $G$ are equivalent if the factors of one are permutation of the factors of another.
Theorem Factors of a composition series are simple.
proof: Let us have a normal series with $\cdots \unlhd H_i \unlhd H_{i+1} \unlhd \cdots$ where $H_{i+1}/H_i$ isn't simple, say it has some normal subgroup $H_i/H_i \lhd N \lhd H_{i+1}/H_i$, then by (Noether4) lattice isomorphism we get the existence of a subgroup $H_i \lhd N' \lhd H_{i+1}$. A refinement of our series.
Thursday 17 January 2013
Inner and Outer automorphisms
Definition The center $Z(G)$ of a group $G$ is the group of elements $z$ such that $\forall g \in G,\,\, zg = gz$. It is abelian and a normal subgroup of $G$.
Definition The set of automorphisms (isomorphisms from $G$ to $G$) $Aut(G)$ is a group under composition.
Notation: We will often write $x^\alpha$ for applying automorphisms $\alpha(x)$ and $x^g = g^{-1} x g$.
Definition The inner automorphisms $\theta_g(x) = g^{-1} x g$ form a group $Inn(G)$. $$(x^{g})^{g'} = \theta_{g'}(\theta_g(x)) = (gg')^{-1} x g g' = \theta_{gg'}(x) = x^{gg'}.$$
Theorem $G/Z(G) \simeq Aut(g)$.
The kernel of the group homomorphism $\theta : G \to Aut(G)$ are the elements $g$ such that $\theta_g(x)=x^g=g^{-1}xg=x$, this is exactly $Z(G)$.
Lemma $N$ is a normal subgroup of $G$ if $gNg^{-1} \subseteq N$.
proof: since group elements are invertible we actually have the equalty $gNg^{-1} = N$ multiply by $g$ to get the original definition of normal.
Theorem $Inn(G) \unlhd Aut(G)$.
Let $\theta_g$ be some inner automorphism then $\forall \alpha$ we have $\alpha (\theta_g (\alpha^{-1}(x))) = \alpha^{-1}(g^{-1})x\alpha^{-1}(g) = \theta_{\alpha(g)}$.
Definition The outer automorphism group is the quotient $Out(G) = Aut(G)/Inn(g)$.
Definition The set of automorphisms (isomorphisms from $G$ to $G$) $Aut(G)$ is a group under composition.
Notation: We will often write $x^\alpha$ for applying automorphisms $\alpha(x)$ and $x^g = g^{-1} x g$.
Definition The inner automorphisms $\theta_g(x) = g^{-1} x g$ form a group $Inn(G)$. $$(x^{g})^{g'} = \theta_{g'}(\theta_g(x)) = (gg')^{-1} x g g' = \theta_{gg'}(x) = x^{gg'}.$$
Theorem $G/Z(G) \simeq Aut(g)$.
The kernel of the group homomorphism $\theta : G \to Aut(G)$ are the elements $g$ such that $\theta_g(x)=x^g=g^{-1}xg=x$, this is exactly $Z(G)$.
Lemma $N$ is a normal subgroup of $G$ if $gNg^{-1} \subseteq N$.
proof: since group elements are invertible we actually have the equalty $gNg^{-1} = N$ multiply by $g$ to get the original definition of normal.
Theorem $Inn(G) \unlhd Aut(G)$.
Let $\theta_g$ be some inner automorphism then $\forall \alpha$ we have $\alpha (\theta_g (\alpha^{-1}(x))) = \alpha^{-1}(g^{-1})x\alpha^{-1}(g) = \theta_{\alpha(g)}$.
Definition The outer automorphism group is the quotient $Out(G) = Aut(G)/Inn(g)$.
Noethers isomorphism theorems
Definition $N$ is a normal subgroup of $G$, written $N\unlhd G$, when $\forall g \in G, \,\, gN = Ng$. Note! This relation is not transitive.
Theorem Let $\varphi : G \to H$ be a group homomorphism, its kernel is a normal subgroup of $G$.
proof: Both $g\ker(\varphi)$ and $\ker(\varphi)g$ are the set of all elements of $G$ which get mapped to $\varphi(g)$.
Definition Let $N \unlhd G$ then $G/N$ is the group of cosets of the form $Ng$.
Lemma $GG = G$. proof: $gG = G$.
Theorem (Noether(-1)) Let $\theta : G \to K$ be a surjective homomorphism and $H \le G$ ($H \unlhd G$) then $\theta(H) \le K$ ($\theta(H) \unlhd K$).
Theorem Let $\varphi : G \to H$ be a group homomorphism, its kernel is a normal subgroup of $G$.
proof: Both $g\ker(\varphi)$ and $\ker(\varphi)g$ are the set of all elements of $G$ which get mapped to $\varphi(g)$.
Definition Let $N \unlhd G$ then $G/N$ is the group of cosets of the form $Ng$.
Lemma $GG = G$. proof: $gG = G$.
Theorem (Noether(-1)) Let $\theta : G \to K$ be a surjective homomorphism and $H \le G$ ($H \unlhd G$) then $\theta(H) \le K$ ($\theta(H) \unlhd K$).
proof: First part is easy, let's just show the bit about normality. $H \unlhd G$ means (equivalent to the definition we gave) $\forall g \in G, h \in H$, $g^{-1}h g \in H$. So let $\forall k \in K, h \in \theta(H)$, write these elements in the form $k = \theta(\bar k)$, $h = \theta(\bar h)$ now $\bar g^{-1}\bar h \bar g \in H$ so applying the homomorphism we get $k^{-1} h k \in \theta(H)$ proving $\theta(H) \unlhd K$.
Theorem (Noether0) Let $N \unlhd G$ then the natural map $x \mapsto Nx : G \mapsto G/N$ is a surjective homomorphism with kernel $N$.
Lemma For $H \le G$ (meaning a subgroup, not necessarily normal) then for $g \in G$, $gH = H$ implies $g \in H$.
proof: Suppose not, then neither is $g^{-1}$ so $1 \not \in H$, contradiction.
Theorem (Noether1a) Let $\varphi$ be surjective, then $H \simeq G/\ker(\varphi)$.
proof: If $g \ker(\varphi) = g' \ker(\varphi)$ then $\varphi(g) = \varphi(g')$ because $g^{-1} g' \in \ker(\varphi)$ so $g \ker(\varphi) \mapsto \varphi(g)$ is well defined and in fact it is a bijection because it's (clearly) surjective on a finite set.
Corollary (Noether1b) Let $\varphi : G \to H$ be any group homomorphism, then $G/\ker(\varphi) \simeq \operatorname{im}(\varphi)$.
Lemma Let $A$ and $B$ be subgroups of $G$ then $A \cap B$ is a subgroup of them both.
Theorem (Noether2) Let $H \le G$ (just a subgroup) and $N \unlhd G$ then $N \cap H \unlhd H$ and $H/(N \cap H) \simeq NH/N$.
proof: Any $x \in H$ may be considered an element of $NH$ since $1 \in N$, so the map $x \mapsto Nx$ is an surjective homomorphism from $H$ to $NH/N$ with kernel $N \cap H$.
Theorem (Noether3) Let $M$ and $N$ be normal subgroups of $G$ and $N \le M$ then $N \unlhd M$, $M/N \unlhd G/N$ and $(G/N)/(M/N) \simeq G/M$.
proof: Define $\alpha : G/N \to G/M$ by $\alpha(Nx) = Mx$ (this is well defined since if $Nx = Nx'$ then by $M \unlhd G$ we get $Mx = Mx'$), the kernel is $M/N$ so this must be a group which implies $N \unlhd M$ because ???. As $\alpha$ is surjective we get the theorem.
Theorem (Noether0) Let $N \unlhd G$ then the natural map $x \mapsto Nx : G \mapsto G/N$ is a surjective homomorphism with kernel $N$.
Lemma For $H \le G$ (meaning a subgroup, not necessarily normal) then for $g \in G$, $gH = H$ implies $g \in H$.
proof: Suppose not, then neither is $g^{-1}$ so $1 \not \in H$, contradiction.
Theorem (Noether1a) Let $\varphi$ be surjective, then $H \simeq G/\ker(\varphi)$.
proof: If $g \ker(\varphi) = g' \ker(\varphi)$ then $\varphi(g) = \varphi(g')$ because $g^{-1} g' \in \ker(\varphi)$ so $g \ker(\varphi) \mapsto \varphi(g)$ is well defined and in fact it is a bijection because it's (clearly) surjective on a finite set.
Corollary (Noether1b) Let $\varphi : G \to H$ be any group homomorphism, then $G/\ker(\varphi) \simeq \operatorname{im}(\varphi)$.
Lemma Let $A$ and $B$ be subgroups of $G$ then $A \cap B$ is a subgroup of them both.
Theorem (Noether2) Let $H \le G$ (just a subgroup) and $N \unlhd G$ then $N \cap H \unlhd H$ and $H/(N \cap H) \simeq NH/N$.
proof: Any $x \in H$ may be considered an element of $NH$ since $1 \in N$, so the map $x \mapsto Nx$ is an surjective homomorphism from $H$ to $NH/N$ with kernel $N \cap H$.
Theorem (Noether3) Let $M$ and $N$ be normal subgroups of $G$ and $N \le M$ then $N \unlhd M$, $M/N \unlhd G/N$ and $(G/N)/(M/N) \simeq G/M$.
proof: Define $\alpha : G/N \to G/M$ by $\alpha(Nx) = Mx$ (this is well defined since if $Nx = Nx'$ then by $M \unlhd G$ we get $Mx = Mx'$), the kernel is $M/N$ so this must be a group which implies $N \unlhd M$ because ???. As $\alpha$ is surjective we get the theorem.
- Nice proofs here: [Ash] Abstract Algebra: The Basic Graduate Year
- Really nice presentation: math.uc.edu/~tsmith/Math610/isothm.pdf
- Loads of properties of normal subgroups: ProofWiki:Normal_Subgroups
- Even better.. GroupProps:Normal_subgroup
first post
This blog will be notes on group theory. You should know a little bit about groups and what they are to read this.
The focus is on finite simple groups. Sometimes definitions and theorems are stated without proof, this means the reader should fill it in. Please tell me if you find mistakes or want to make suggestions.
$$G \to G/[G,G]$$
Not underlined means it's (should be..) "revision"
Section 1:
The focus is on finite simple groups. Sometimes definitions and theorems are stated without proof, this means the reader should fill it in. Please tell me if you find mistakes or want to make suggestions.
$$G \to G/[G,G]$$
- Amazingly good intro group theory: GroupProps: Getting Started
- McKay's great proof of Cauchy's theorem
Not underlined means it's (should be..) "revision"
Section 1:
- Noethers Isomorphism theorems.
- Automorphisms (Inner and Outer).
- Group actions and the orbit-stabilizer theorem.
- Normalizers and the conjugacy-class equation
- Sylow p-groups (uses orbits)
- Commutator subgroups.
- Series. Composition Series. Central Series. Jordan-Holder.
- ABC.
- Optimal series for solvable groups
- Optimal series for nilpotent groups
- Structure of Nilpotent: product of p-groups
- Halls theorem
- Structure of Solvable: Sylow-bases
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