Theorem (Jordan-Holder) Any two composition series are equivalent.
proof: Induction of the length of the series. Suppose we have 1 \lhd \cdots \lhd L \lhd G and 1 \lhd \cdots \lhd K \lhd G then L \lhd KL \lhd G implies that KL/L is a normal subgroup of the composition factor G/L which being a simple group implies that KL = L\text{ or }G and similarly KL = K\text{ or }G.
Suppose KL \not = G, then L = K and we are done by induction. Suppose KL = G then by noether2 we have \frac{G}{L} \simeq \frac{KL}{L} \simeq \frac{K}{K \cap L} and \frac{G}{K} \simeq \frac{KL}{K} \simeq \frac{L}{K \cap L} therefore the composition series
\begin{array}{c}
1 &\lhd& \cdots &\lhd& L \lhd G \\
1 &\lhd& \cdots \lhd L \cap K &\lhd& L \lhd G \\
1 &\lhd& \cdots \lhd L \cap K &\lhd& K \lhd G \\
1 &\lhd& \cdots &\lhd& K \lhd G \\
\end{array} are equivalent.
finite simple blog
Tuesday, 2 April 2013
Sunday, 10 March 2013
A_n is the only normal subgroup of S_n
Lemma Suppose 1 \not = N \lhd G has trivial intersection with [G,G], then it lies in the center.
proof: Let n \in N then n^g [g,n] = g^{-1} n g [g,n] = g^{-1} g n = n and we know n^g \in N so [g,n] \in N so it equals 1 so n commutes with g.
Lemma S_n for n \ge 3 has trivial center.
proof: If z lies in the center then zg = gz for all \pi. We show that g^z = g for all g implies z=1: Take any three symbols from the group a,b,c then consider:
Note: S_4 has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).
proof: Let n \in N then n^g [g,n] = g^{-1} n g [g,n] = g^{-1} g n = n and we know n^g \in N so [g,n] \in N so it equals 1 so n commutes with g.
Lemma S_n for n \ge 3 has trivial center.
proof: If z lies in the center then zg = gz for all \pi. We show that g^z = g for all g implies z=1: Take any three symbols from the group a,b,c then consider:
- (a\;b)^z = (az\;bz) so az=a,bz=b or az=b,bz=a.
- (a\;b\;c)^z = (az\;bz\;cz) so (using the previous) cz=c.
Note: S_4 has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).
Tuesday, 5 March 2013
Outer Automorphism of S_6
Lemma If \alpha \in Aut(S_n) maps transpositions to transpositions iff it's an inner automorphism.
proof: (\Leftarrow) inner automorphisms are done by conjugation which preserves cycle type. (\Rightarrow) todo
Lemma If n \not = 6 then Out(S_n) = 1.
proof: An outer-automorphism must swap transpositions with some other order-2 conjugacy class. First we count cycles of type 2^k 1^{n-2k}. You have \binom{n}{n-2k} choices of fixed elements for each, then with the 2k remaining elements we can permute these in 2k! ways before factoring out the order each transposition is written in and the number of ways we order the transpositions 2^k k!, therefore there are f_k^n = \frac{n!}{(2k)! (n-2k)!} \cdot \frac{(2k)!}{2^k k!} cycles of the given type. In particular f_1^n = \binom{n}{2}.
For n > 6, k > 1 f_k^n > \binom{n}{2k} \ge \binom{n}{2} unless n = 2k or 2k+1 - but in those cases prove these cases cannot occur - so there can be no outer automorphisms. For n < 6 do this too.
Theorem S_6 has an outer-automorphism.
proof: There are 6 S_5 subgroups as point stabilizers, but from the following diagram we find an S_5 that is not a point stabilizer
Permutations of the 5 colors correspond to permutations of the 6 points (which we label 1,2,3,4,5,6 clockwise starting at the top left) (\color{red}{\text{red}}\;\color{yellow}{\text{yellow}})=(1\;2)(3\;6)(4\;5) (\color{blue}{\text{blue}}\;\color{green}{\text{green}})(\color{red}{\text{red}}\;\color{purple}{\text{purple}}\;\color{yellow}{\text{yellow}})=(1\;2\;3\;4\;5\;6).
So we have discovered an exotic S_5 inside S_6, I do not know why but there are 6 conjugates of it. We will call the action of S_6 on these 6 conjugates \varsigma.
Note, that (1\;2)\varsigma = (1\;2)(3\;4)(5\;6) means we have an outer automorphism! well, if we have an automorphism:
and we do.
proof: (\Leftarrow) inner automorphisms are done by conjugation which preserves cycle type. (\Rightarrow) todo
Lemma If n \not = 6 then Out(S_n) = 1.
proof: An outer-automorphism must swap transpositions with some other order-2 conjugacy class. First we count cycles of type 2^k 1^{n-2k}. You have \binom{n}{n-2k} choices of fixed elements for each, then with the 2k remaining elements we can permute these in 2k! ways before factoring out the order each transposition is written in and the number of ways we order the transpositions 2^k k!, therefore there are f_k^n = \frac{n!}{(2k)! (n-2k)!} \cdot \frac{(2k)!}{2^k k!} cycles of the given type. In particular f_1^n = \binom{n}{2}.
For n > 6, k > 1 f_k^n > \binom{n}{2k} \ge \binom{n}{2} unless n = 2k or 2k+1 - but in those cases prove these cases cannot occur - so there can be no outer automorphisms. For n < 6 do this too.
Theorem S_6 has an outer-automorphism.
proof: There are 6 S_5 subgroups as point stabilizers, but from the following diagram we find an S_5 that is not a point stabilizer
Permutations of the 5 colors correspond to permutations of the 6 points (which we label 1,2,3,4,5,6 clockwise starting at the top left) (\color{red}{\text{red}}\;\color{yellow}{\text{yellow}})=(1\;2)(3\;6)(4\;5) (\color{blue}{\text{blue}}\;\color{green}{\text{green}})(\color{red}{\text{red}}\;\color{purple}{\text{purple}}\;\color{yellow}{\text{yellow}})=(1\;2\;3\;4\;5\;6).
So we have discovered an exotic S_5 inside S_6, I do not know why but there are 6 conjugates of it. We will call the action of S_6 on these 6 conjugates \varsigma.
gap> s5 := Group((1,2)(3,6)(4,5),(1,3,6,5,4)); Group([ (1,2)(3,6)(4,5), (1,3,6,5,4) ]) gap> ex := ConjugateSubgroups(SymmetricGroup(6),s5); [ Group([ (1,2)(3,6)(4,5), (1,3,6,5,4) ]), Group([ (1,2)(3,5)(4,6), (1,3,5,6,4) ]), Group([ (1,2)(3,6)(4,5), (1,3,6,4,5) ]), Group([ (1,2)(3,4)(5,6), (1,3,4,6,5) ]), Group([ (1,2)(3,5)(4,6), (1,3,5,4,6) ]), Group([ (1,2)(3,4)(5,6), (1,3,4,5,6) ]) ] gap> Position(ex, ex[1]^(1,2)); 2 gap> Position(ex, ex[2]^(1,2)); 1 gap> Position(ex, ex[3]^(1,2)); 4 gap> Position(ex, ex[4]^(1,2)); 3 gap> Position(ex, ex[5]^(1,2)); 6 gap> Position(ex, ex[6]^(1,2)); 5
Note, that (1\;2)\varsigma = (1\;2)(3\;4)(5\;6) means we have an outer automorphism! well, if we have an automorphism:
gap> Position(ex, ex[1]^(1,2,3,4,5,6)); 1 gap> Position(ex, ex[2]^(1,2,3,4,5,6)); 3 gap> Position(ex, ex[3]^(1,2,3,4,5,6)); 2 gap> Position(ex, ex[4]^(1,2,3,4,5,6)); 5 gap> Position(ex, ex[5]^(1,2,3,4,5,6)); 6 gap> Position(ex, ex[6]^(1,2,3,4,5,6)); 4 gap> StructureDescription(Group((1,2)(3,4)(5,6),(2,3)(4,5,6))); "S6"
and we do.
Friday, 1 March 2013
The first 3 p-groups
Definition The equivalence relation a \sim b iff \exists g, a^g = b partitions a group into conjugacy classes. Each elements of the center of a group is its own conjugacy class.
Proposition The number of cosets of a centralizers is the same as the number of elements of a conjugacy classes.
proof: Let C be a conjugacy class so that C^g = C for all g. Let a \in C then the orbit of a under the conjugacy action fills up the whole of C. The stabilizer of this action is equal to the centralizer of a so by orb-stab we have |C| = |G:C_G(a)|.
Lemma Prime power order implies not centerless.
proof: Let G act on itself by conjugation, clearly Z(G) is invariant with respect to this action. We have the conjugacy class equation, where the sum runs over conjugacy class representatives |G| = |Z(G)| + \sum_{g}|G:C_G(g)| with C_G(g) = \{x \in G\mid \forall x \in G, xg = gx \} being the centralizer of g. Using the fact that C_G(g) is never the whole group (otherwise g commutes with everything, so it would be in the center instead) we deduce the lemma \mod p.
Lemma A nonabelian group can never have a nontrivial cyclic quotient.
proof: Let G/N be generated by gN so that every coset is of the form g^i N and so every element of the group is of the form g^i n. Then the group is abelian since g^i n g^j n' = g^{i+j} n n' = g^j n' g^i n.
Theorem |G|=p then G=C_p.
proof: Cauchy's theorem gives an element of order p, it must generate the whole group.
Theorem |G|=p^2 then G=C_{p^2} or C_p^2.
proof: We know from counting conjugacy classes that |Z(G)| is p or p^2 and it can't be p by the lemma (because |G/Z(G)|=p), so the group is abelian.
Theorem |G|=p^3 then G is abelian or |Z(G)|=p
proof: In the non-abelian case |Z(G)| must be p or p^2, but p^2 cannot occur by the lemma since then |G/Z(G)|=p would be cyclic.
Classification The nonabelian groups of order 2^3 are D_8 and Q.
Proposition The number of cosets of a centralizers is the same as the number of elements of a conjugacy classes.
proof: Let C be a conjugacy class so that C^g = C for all g. Let a \in C then the orbit of a under the conjugacy action fills up the whole of C. The stabilizer of this action is equal to the centralizer of a so by orb-stab we have |C| = |G:C_G(a)|.
Lemma Prime power order implies not centerless.
proof: Let G act on itself by conjugation, clearly Z(G) is invariant with respect to this action. We have the conjugacy class equation, where the sum runs over conjugacy class representatives |G| = |Z(G)| + \sum_{g}|G:C_G(g)| with C_G(g) = \{x \in G\mid \forall x \in G, xg = gx \} being the centralizer of g. Using the fact that C_G(g) is never the whole group (otherwise g commutes with everything, so it would be in the center instead) we deduce the lemma \mod p.
Lemma A nonabelian group can never have a nontrivial cyclic quotient.
proof: Let G/N be generated by gN so that every coset is of the form g^i N and so every element of the group is of the form g^i n. Then the group is abelian since g^i n g^j n' = g^{i+j} n n' = g^j n' g^i n.
Theorem |G|=p then G=C_p.
proof: Cauchy's theorem gives an element of order p, it must generate the whole group.
Theorem |G|=p^2 then G=C_{p^2} or C_p^2.
proof: We know from counting conjugacy classes that |Z(G)| is p or p^2 and it can't be p by the lemma (because |G/Z(G)|=p), so the group is abelian.
Theorem |G|=p^3 then G is abelian or |Z(G)|=p
proof: In the non-abelian case |Z(G)| must be p or p^2, but p^2 cannot occur by the lemma since then |G/Z(G)|=p would be cyclic.
Classification The nonabelian groups of order 2^3 are D_8 and Q.
Thursday, 28 February 2013
Simplicity of the projective special linear groups
Theorem If n \ge 2 then PSL_n(q) is simple provided (n,q) is not (2,2) or (2,3).
proof: Consider PSL_n(q) acting on \mathbb P^{n-1}(q) for n \ge 2 and the exceptions do not occur, then it is primitive since it's 2-transitive and by previous results SL_n(q) is perfect, so since PSL_n(q) is a quotient of that it's perfect too. Take d \in V^\# so that [d] \in \mathbb P^{n-1}(q) let A be the image of \mathscr T(d) in PSL_n(q) applying [???] and taking quotients we see that A is a normal abelian subgroup of the stabilizer PSL_n(q)_{[d]} and it's conjugates generate PSL_n(q) thus the conditions of Iwasawa's lemma are satisfied.
proof: Consider PSL_n(q) acting on \mathbb P^{n-1}(q) for n \ge 2 and the exceptions do not occur, then it is primitive since it's 2-transitive and by previous results SL_n(q) is perfect, so since PSL_n(q) is a quotient of that it's perfect too. Take d \in V^\# so that [d] \in \mathbb P^{n-1}(q) let A be the image of \mathscr T(d) in PSL_n(q) applying [???] and taking quotients we see that A is a normal abelian subgroup of the stabilizer PSL_n(q)_{[d]} and it's conjugates generate PSL_n(q) thus the conditions of Iwasawa's lemma are satisfied.
Wednesday, 27 February 2013
M11
Let \Delta = \{\delta_1, \delta_2, \delta_3, \delta_4, \delta_5\} and let \Omega be the unordered pairs of these. We will use the names:
The stabilizer of 0 has size |G_0| = |G|/|\Omega| = 60/10 = 6 (orbit stabilizer) so G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\} and the G_0 orbits are:
so take \infty \not\in \Omega and set \Omega^+ = \Omega \cup \{\infty\} then choose x = (\infty\;0)(2\;3)(4\;6)(8\;9) then
Theorem L is simple.
1 and x are (G,G)-double coset reps in L. Take \omega \not \in \Omega^+ and form \Omega^\star = \Omega^+ \cup \{\omega\}. y=(\omega\;\infty)(1\;4)(2\;5)(3\;6) then
Theorem M_{11} is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
- 0: \{\delta_1,\delta_2\}, 1: \{\delta_1,\delta_3\}, 2: \{\delta_1,\delta_4\}
- 3: \{\delta_1,\delta_5\}, 4: \{\delta_2,\delta_3\}, 5: \{\delta_2,\delta_4\}
- 6: \{\delta_2,\delta_5\}, 7: \{\delta_3,\delta_4\}, 8: \{\delta_3,\delta_5\}
- 9: \{\delta_4,\delta_5\}
- a = (\delta_1\;\delta_2)(\delta_3\;\delta_5) = (1\;6)(2\;5)(3\;4)(7\;9)
- b = (\delta_1\;\delta_2)(\delta_4\;\delta_5) = (1\;4)(2\;6)(3\;5)(7\;8)
- g_2 = (\delta_2\;\delta_3)(\delta_4\;\delta_5) = (0\;1)(2\;3)(5\;8)(6\;7)
- g_3 = (\delta_1\;\delta_4)(\delta_2\;\delta_3) = (0\;7)(1\;5)(3\;9)(6\;8)
The stabilizer of 0 has size |G_0| = |G|/|\Omega| = 60/10 = 6 (orbit stabilizer) so G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\} and the G_0 orbits are:
- \{0\}
- \{1,2,3,4,5,6\}
- \{7,8,9\}
gap> a:=(1,2)(3,5);;b:=(1,2)(4,5);; gap> DoubleCosets(Group((1,2,3,4,5),(1,2,3)),Group(a,b),Group(a,b)); [ DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(),Group( [ (1,2)(3,5), (1,2)(4,5) ] )), DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(2,3)(4,5),Group( [ (1,2)(3,5), (1,2)(4,5) ] )), DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(1,3)(2,4),Group( [ (1,2)(3,5), (1,2)(4,5) ] )) ]
so take \infty \not\in \Omega and set \Omega^+ = \Omega \cup \{\infty\} then choose x = (\infty\;0)(2\;3)(4\;6)(8\;9) then
- x^2 = 1 \in G_0
- a^x = b and (since x has order 2) b^x = a so G_0^x = G_0.
- g_2^x = (\infty\;1)(2\;3)(5\;9)(4\;7)=x^{g_2} \in G x G.
- g_3^x = (\infty\;7)(1\;5)(2\;8)(4\;9)=x^{g_3} a^b \in G x G.
Theorem L is simple.
1 and x are (G,G)-double coset reps in L. Take \omega \not \in \Omega^+ and form \Omega^\star = \Omega^+ \cup \{\omega\}. y=(\omega\;\infty)(1\;4)(2\;5)(3\;6) then
- y^2 = 1 \in G = L_{\infty}
- G^y = G since G = \langle a,b,g_2,g_3 \rangle and a^y = a, b^y = b, g_2^y = g_2^b, g_3^y = g_3 a^b.
- x^y = y^x a^b \in L y L
Theorem M_{11} is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
A_n is simple!
Lemma A_5 is perfect.
proof: A_5 is generated by (1\;2\;3) and (1\;2\;3\;4\;5) both are commutators:
Theorem A_5 is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides |A_5| = 60 is solvable so A_5 has no normal subgroups (else it would be solvable too!)
Theorem A_n is simple.
proof: Induction on n with base case 5. A_n is n-2 transitive in the natural action (by the multiple-transitivity section) so for n > 5 this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be C_2^2 in the case of A_6 and there isn't one otherwise but C_2^2 doesn't have enough elements to be transitive on 6 points so it can't be regular - so A_n has no regular normal subgroups: therefore by the proposition in that section it's simple.
proof: A_5 is generated by (1\;2\;3) and (1\;2\;3\;4\;5) both are commutators:
gap> a := (1,5,2);; b := (4,2,3);; a^(-1)*b^(-1)*a*b; (1,2,3) gap> a := (1,2,3)*(3,4,5);; b := (1,4,2)*(3,5,2);; a^(-1)*b^(-1)*a*b; (1,2,3,4,5)
Theorem A_5 is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides |A_5| = 60 is solvable so A_5 has no normal subgroups (else it would be solvable too!)
gap> List(AllSmallGroups(2), StructureDescription); [ "C2" ] gap> List(AllSmallGroups(2^2), StructureDescription); [ "C4", "C2 x C2" ] gap> List(AllSmallGroups(2*3), StructureDescription); [ "S3", "C6" ] gap> List(AllSmallGroups(2*5), StructureDescription); [ "D10", "C10" ] gap> List(AllSmallGroups(2*3*5), StructureDescription); [ "C5 x S3", "C3 x D10", "D30", "C30" ] gap> List(AllSmallGroups(2^2*3), StructureDescription); [ "C3 : C4", "C12", "A4", "D12", "C6 x C2" ] gap> List(AllSmallGroups(2^2*5), StructureDescription); [ "C5 : C4", "C20", "C5 : C4", "D20", "C10 x C2" ]
Theorem A_n is simple.
proof: Induction on n with base case 5. A_n is n-2 transitive in the natural action (by the multiple-transitivity section) so for n > 5 this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be C_2^2 in the case of A_6 and there isn't one otherwise but C_2^2 doesn't have enough elements to be transitive on 6 points so it can't be regular - so A_n has no regular normal subgroups: therefore by the proposition in that section it's simple.
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