Normal Subgroups, Quotient Groups, and Automorphisms

December 10, 2025 · Luciano Muratore

So far we have seen how groups can be generated and how they can be combined. Now we turn to a more structural question: what subgroups of a group are well-behaved enough to let us form a new group out of their cosets?

Normal Subgroups

A subgroup $H \leq G$ is called normal if it is invariant under conjugation by any element of $G$, that is, if $gHg^{-1} = H$ for all $g \in G$, or equivalently, if left and right cosets coincide: $gH = Hg$ for all $g \in G$.

There are several ways to recognize normality. One is that $H$ is a union of conjugacy classes. Another is that $xHx^{-1} \subseteq H$ for all $x \in G$ — which, combined with the finite or symmetric argument, gives equality. And perhaps the most important one: $H$ is normal if and only if the cosets form a group $G/H$, the quotient group.

A simple but useful criterion is the following: if $[G : H] = 2$, then $H$ is normal. The reason is purely combinatorial — there are only two left cosets ($H$ and $gH$) and only two right cosets ($H$ and $Hg$), so for $g \notin H$, both $gH$ and $Hg$ must equal $G \setminus H$, forcing $gH = Hg$. The canonical example is $A_n \trianglelefteq S_n$, since $[S_n : A_n] = 2$.

Normal Subgroups of $S_3$

Let us work through a concrete case. The symmetric group $S_3 = \{e, (12), (13), (23), (123), (132)\}$ has subgroups $\{e\}$, $\langle(12)\rangle$, $\langle(13)\rangle$, $\langle(23)\rangle$, $A_3$, and $S_3$ itself. Of these, only $\{e\}$, $A_3$, and $S_3$ are normal.

Why are the order-2 subgroups not normal? Because conjugation sends one transposition to another — for example, $(123)(12)(132) = (23)$ — so $\langle(12)\rangle$ is not preserved by all conjugations. On the other hand, $A_3$ has index 2 and is therefore normal by the criterion above, and $S_3/A_3 \cong \mathbb{Z}_2$.

The Quaternion Group $Q_8$

The quaternion group $Q_8 = \{1, -1, i, -i, j, -j, k, -k\}$ with relations $i^2 = j^2 = k^2 = -1$, $ij = k$, $ji = -k$ is another instructive example. Its center is $Z(Q_8) = \{1, -1\}$, and its subgroups are $\{1\}$, $\{1,-1\}$, $\langle i \rangle$, $\langle j \rangle$, $\langle k \rangle$, and $Q_8$.

Remarkably, every subgroup of $Q_8$ is normal. The reason is that conjugation sends any generator to itself or its inverse: for example, $j \cdot i \cdot j^{-1} = j \cdot i \cdot (-j) = -ji \cdot j^{-1}$… and one can check directly that each subgroup is preserved. The quotient groups are also illuminating: the index-2 quotients $Q_8/\langle i \rangle$, $Q_8/\langle j \rangle$, $Q_8/\langle k \rangle$ are all isomorphic to $\mathbb{Z}_2$, and the index-4 quotient $Q_8/\{1,-1\} \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

Products of Normal Subgroups

If $H, K \trianglelefteq G$, then their intersection $H \cap K$ is normal in $G$, and their product $HK = \{hk \mid h \in H, k \in K\}$ equals $\langle H \cup K \rangle$ and is also normal in $G$. The key idea is that normality ensures conjugation preserves both $H$ and $K$, so it preserves their product as well.

Homomorphisms

A map $f : G \to H$ is a homomorphism if $f(xy) = f(x)f(y)$ for all $x, y \in G$. The kernel $\ker(f) = \{x \mid f(x) = e_H\}$ is always a normal subgroup of $G$, and the image $\text{Im}(f) = f(G)$ is a subgroup of $H$.

Some examples worth keeping in mind: $f(x) = e^x$ is a homomorphism from $(\mathbb{R}, +)$ to $(\mathbb{R}_{>0}, \cdot)$; $f(x) = e^{ix}$ maps $\mathbb{R}$ to the unit circle; and $f(x,y) = x$ is a projection from $\mathbb{Z}^2$ to $\mathbb{Z}$.

Endomorphisms and Automorphisms of $\mathbb{Z}$

Every endomorphism $\varphi : \mathbb{Z} \to \mathbb{Z}$ is determined by $\varphi(1) = k$, giving $\varphi(n) = kn$. For $\varphi$ to be an automorphism it must be bijective, which forces $|k| = 1$. So $\text{Aut}(\mathbb{Z}) = \{\varphi_1, \varphi_{-1}\} \cong \mathbb{Z}_2$: the only automorphisms of $\mathbb{Z}$ are the identity and the inversion map $n \mapsto -n$.

Inner Automorphisms

For any $g \in G$, conjugation by $g$ defines an automorphism $\sigma_g : G \to G$ by $\sigma_g(x) = gxg^{-1}$. These satisfy $\sigma_g \circ \sigma_h = \sigma_{gh}$ and $(\sigma_g)^{-1} = \sigma_{g^{-1}}$, so the collection $\text{Int}(G) = \{\sigma_g \mid g \in G\}$ forms a subgroup of $\text{Aut}(G)$, and in fact a normal one.

Now consider the map $f : G \to \text{Int}(G)$ defined by $f(g) = \sigma_g$. It is a surjective homomorphism, and its kernel is precisely $Z(G)$, the center: $g \in \ker(f)$ if and only if $gxg^{-1} = x$ for all $x$, that is, if and only if $g$ commutes with everything. By the First Isomorphism Theorem,

$G/Z(G) \cong \text{Int}(G).$

This is a neat result: the group of inner automorphisms is completely determined by how far $G$ is from being abelian.

The Inversion Map

Let me close with a small observation that ties back to what we saw about groups of exponent 2. Consider the map $f(x) = x^{-1}$. When is this a homomorphism?

For $f$ to satisfy $f(xy) = f(x)f(y)$, we would need $(xy)^{-1} = x^{-1}y^{-1}$. But $(xy)^{-1} = y^{-1}x^{-1}$, so the condition becomes $y^{-1}x^{-1} = x^{-1}y^{-1}$, which is exactly the statement that $G$ is abelian. So $f$ is a homomorphism if and only if $G$ is abelian — a compact way of saying that inversion reverses order, and the group only forgives that if order does not matter.