# Homomorphisms and Kernels

Homomorphisms are more general than isomorphisms, except that $f:G\to H$ need not be a bijection. For example, suppose $G=\mathbf{Z}$ and $H=\mathbf{Z}/n\mathbf{Z}$. Then define $f:G\to H$ by $f:a\mapsto a\bmod n$. This is not injective, because $k,k+n,k+2n,\cdots$ all go to $k\bmod n$. It’s a homomorphism.

Lemma: Suppose $f:G\to H$ is a homomorphism. Then $f(e)=e$$f(a)^{-1}=f(a^{-1})$, and if $K\leq G$, $f(K)\leq H$.

Proof: First: If $a=f(e)$, then $a=f(ee)=f(e)f(e)=f(e)$, so $f(e)$ is its own inverse, and $f(e)=e$.
Second: $e=f(e)=f(aa^{-1})=f(a)f(a^{-1})$, so $f(a^{-1})=f(a)^{-1}$.
Third: We want to prove that it’s closed under multiplication and inverses. Suppose $g,h\in f(K)$. Then $g=f(a),h=f(b)$ for some $a,b\in G$. Thus $gh=f(a)f(b)=f(ab)$ so $gh\in f(K)$. This shows that it’s closed under multiplication. Suppose $g\in f(K)$. Then $g=f(a)$ for some $a\in G$. Now $g^{-1}=f(a)^{-1}=f(a^{-1})$ so $g^{-1}\in f(K)$, i.e., it’s closed under inverses. QED.

Definition: Suppose $K\leq G$. Then $K$ is normal if $gKg^{-1}\subseteq K$ for all $g\in G$.

This implies that $gH=Hg$, and actually, $gHg^{-1}=H$. If $G$ is abelian, i.e., commutative, then any subgroup of $G$ is normal.

Definition: Let $f:G\to H$ be a homomorphism. Then define $\ker f=\{g\in G|f(g)=e\}$. This is called the kernel of $f$.

Proposition: Let $f:G\to H$ be a homomorphism. Then $\ker f=\{e\}$ if and only if $f$ is injective.

Proof: Suppose $f$ is injective. Then suppose $\ker f=\{g_i\}_{i\in I}$ (i.e. $\ker f$ is a collection of elements of $G$ which is indexed by some set $I$). This means that $f(g_i)=f(e)=e$. But $f$ is injective. Thus $g_i=e$; since $i$ is arbitrary, this means that $\ker f=\{e\}$.

Now suppose $\ker f=\{e\}$, and suppose $f(g)=f(h)$. We want to prove that $g=h$ (because this is the definition of an injective function). If $f(g)=f(h)$, then $f(g)f(h^{-1})=f(h)f(h^{-1})=f(h)f(h)^{-1}=e$. So $f(gh^{-1})=\{e\}$. Thus $gh^{-1}\in\ker f$; but $gh^{-1}=e$, and so $g=h$. QED.

Corollary: Suppose $f$ is an isomorphism. Then $\ker f=\{e\}$.

What was the point of introducing the notion of a normal subgroup before, though?

Theorem: Let $f:G\to H$ be a homomorphism. Then $\ker f$ is a normal subgroup of $G$.

Proof: We first need to show that $\ker f$ is a subgroup of $G$. It suffices to prove that it is closed under multiplication and inverses. For multiplication: suppose $g,h\in\ker f$. Then $f(gh)=f(g)f(h)=ee=e$, so $gh\in \ker f$. For inverses: suppose $g\in \ker f$. Then $f(gg^{-1})=f(g)f(g^{-1})=ef(g^{-1})=f(g^{-1})=e$, so $g^{-1}\in\ker f$. So $\ker f$ is a subgroup of $G$.

Now to prove normality. Suppose $g\in G$. We need to show that if $h\in\ker f$, then $ghg^{-1}\in\ker f$. Well, $f(ghg^{-1})=f(g)f(h)f(g^{-1})=f(g)f(h)f(g)^{-1}$, which is equal to $f(g)ef(g)^{-1}=f(g)f(g)^{-1}=e$, so $ghg^{-1}\in\ker f$ for any $g\in G$ and $h\in \ker f$. QED.

We can ask: if $f:G\to H$ is a homomorphism, is $f(G)$ isomorphic to a subgroup of $G$? In general, it can’t be the whole group $G$ itself, because the kernel of $f$ could be nontrivial, i.e., more than one element could map to the identity. (Unless the homomorphism is injective, because then $\ker f$ is trivial.) It turns out that the subgroup of $G$ which $\ker f$ is isomorphic to is something called the quotient group of $G$. (The thing it’s isomorphic to is called the quotient group, but the reason it’s isomorphic to the quotient group is the content of Talk 8, namely the isomorphism theorems. These are the most important theorems in group theory.)

Another, perhaps more simpler way of motivating quotient groups is as follows. Suppose $G$ is a group, and $H\leq G$. Can we find a group $K$ and a homomorphism $\phi:G\to K$ such that $\ker\phi$ is isomorphic to $H$? (Because $\ker\phi$ is normal, you obviously need $H$ to be a normal subgroup of $G$.) Defining the quotient group will be the topic of the next post.

Have fun,
SKD