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
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