While isomorphisms need to be bijections, Homomorphisms do not.

Definition 2.4.1
A map between groups \(\varphi : G \rightarrow H\) is called a homomorphism if it preserves group multiplication, \(\varphi(g_1)(g_2) = \varphi(g_1)\varphi(g_2)\) for all \(g_1, g_2 \in G\). An endomorphism of \(G\) is a homomorphism \(\varphi: G \rightarrow G\).


So Given a homomorphism, if we find that it’s a bijection, then it is an isomorphism.

Some Facts about Homomorphisms

Proposition 2.4.11
Let \(\varphi : G \rightarrow H\) and \(\psi : G \rightarrow H\) be homomorphisms of groups.
  1. \(\varphi(e_G) = e_H\)
  2. For each \(g \in G\), \(\varphi(g^{-1}) = (\varphi(g))^{-1}\)


Proof (Book)
For any \(g \in G\).

$$ \begin{align*} \varphi(e_G)\varphi(g) &= \varphi(e_Gg) \quad \text{(because $\varphi$ is a homomorphism)} \\ &= \varphi(g) \end{align*} $$

So \(\varphi(e_G)\) is an identity element in \(H\) but by the uniqueness of the identity element, then \(\varphi(e_G) = e_H\). Similarly, for any \(\in G\),

$$ \begin{align*} \varphi(g^{-1})\varphi(g) &= \varphi(g^{-1}g) \quad \text{(because $\varphi$ is a homomorphism)} \\ &= \varphi(e_G) \\ & = e_H \end{align*} $$

And by the uniqueness of the inverse, \(\varphi(g^{-1}) = \varphi(g)^{-1}\). \(\ \blacksquare\)


The Image of a Homomorphism

Given a homomorphism \(\varphi\) between two groups \(G\) and \(H\), the image of \(\varphi\) which is a subset of \(H\) is also a subgroup.

Proposition
Given some homomorphism \(\varphi: G \rightarrow H\). The image of this function $$ \begin{align*} \varphi(G) = \{\varphi(g) \ : \ g \in G\} \end{align*} $$ is a subgroup of \(H\).


Proof
TODO


The Kernel of a Homomorphism

The kernel of a group is simply all the elements such that \(\varphi(g)=e\).

Definition 2.4.14
Let \(\varphi: G \rightarrow H\) be a homomorphism of groups. The kernel of the homomorphism \(\varphi\), denoted \(ker(\varphi)\), is \(\varphi^{-1}(e_H) = \{g \in G \ : \ \varphi(g) = e_H\}\).


In fact, The kernel of a group \(K = ker \ \varphi\) is a subgroup of \(G\).

Proof
We will show this by showing that the kernel satisfies the subgroup properties:

  1. We know by Proposition 2.4.11, that \(\varphi(e_G) = e_H\) so \(e_G \in K\).
  2. Closed under product: For any \(a,b \in K\), \(\varphi(a)\varphi(b) = e_He_H = e_H \in K\)
  3. Closed under inverses: For any \(a \in K\), \(\varphi(a^{-1}) = \varphi(a)^{-1} = e_H^{-1} = e_H\). Therefore,\(a^{-1} \in K\). \(\ \blacksquare\)



Examples

Example 1: Consider the following with the addition operation

$$ \begin{align*} \pi \ : \ &\mathbf{Z} \rightarrow \mathbf{Z}_n\\ &a \rightarrow [a]_n \end{align*} $$

\(\pi\) is a homomorphism because \([a+b] = [a] + [b]\). Remember the rule is that \(\varphi(ab) = \varphi(a)\varphi(b)\) where the first operation comes from the first group while the second operation comes from the second group. Here, both groups have addition as their binary operation.

\(\pi\) is also surjective. Since for any element \(h \in \mathbf{Z}_n\), we have an element \(g \in \mathbf{Z}\) such that \(\pi(g) = [g]\). Moreover, we see that \(ker(\pi) = \mathbf{Z}n = \langle n \rangle\).

Example 2: Suppose \(g \in G\). Let

$$ \begin{align*} \varphi \ : \ &\mathbf{Z} \rightarrow G\\ &k \rightarrow g^k \end{align*} $$

We want to show that \(\varphi(ab) = \varphi(a)\varphi(b)\). But we know that \(\varphi(ab) = g^{a+b} = g^ag^b = \varphi(a)\varphi(b)\). So \(\varphi\) is a homomorphism.

Note here that the image of \(\varphi\) is the subgroup generated by \(g\), \(\langle g \rangle\). What about the kernel of \(\varphi\)?

$$ \begin{equation*} ker(\varphi) = \begin{cases} \{0\} \quad &\text{if } o(g) = \infty \\ \mathbf{Z}d \quad \quad &\text{if } o(g) = d \end{cases} \end{equation*} $$


Example 3: Consider the following

$$ \begin{align*} \varphi \ : \ &D_n \rightarrow S_n\\ &V \rightarrow Sym(V) \end{align*} $$

where \(V\) is the vertices of the polygon. In fact \(\varphi(D_n)\) is isomorphic to \(D_n\) and it is a subgroup of \(S_n\).


Example 4: Consider the following

$$ \begin{align*} \varphi \ : \ &S_n \rightarrow GL(\mathbf{R}^{x})\\ &\sigma \rightarrow P_{\sigma} \end{align*} $$

where \(P_{\sigma}\) is the matrix associated with the permutation \(\sigma\). For example

$$ \begin{align*} (1 \ 2 \ 3) \rightarrow \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{align*} $$

Example 5: Consider the following

$$ \begin{align*} \psi \ : \ &GL(\mathbf{R}^{x}) \rightarrow \mathbf{R}^{x}\\ &P \rightarrow \det(P) \end{align*} $$

This is another homomorphism. In fact, The composition of \(\psi \circ \varphi\) is a homomorphism. That is

$$ \begin{align*} \psi \circ \varphi \ : \ &S_n \rightarrow \mathbf{R}^{x}\\ &\sigma \rightarrow \det(P_{\sigma}) \end{align*} $$

is a homomorphism. Fact: The composition of two homomorphism is a homomorphism.

Example 6: Consider \(V\) where \(V\) is a vector space. Now consider the group \((V, +)\) (throw the scalar multiplication away). This is an abelian group. Consider the linear map

$$ \begin{align*} T \ : \ &V \rightarrow W \end{align*} $$

This is a homomorphism (recall that linear maps need to also to preserve scalar multiplication). Furthermore, \(ker T = \{v \in V \ | \ T(v) = 0\}\) (observe that the kernel is also the nullspace of \(T\)).


Normal Subgroups

It turns out that only a certain kind of subgroups could be kernels of homomorphism.

Definition
A normal subgroup of \(G\) is a subgroup \(N \leq G\) such that \(gNg^{-1} = N\) for all \(g \in G\) where \(gNg^{-1} = \{gng^{-1} \ | \ n \in N\}\).


This operation \(x \rightarrow gxg^{-1}\) is called the conjugation of \(x\) by \(g\). The definition says that if we conjeguate all the elements of \(N\) by a fixed element \(g \in G\), then we should get \(N\) back for every \(g\).

Proposition
To show a subgroup \(N \leq G\) is normal, it suffies to show that \(gNg^{-1} \subseteq N\) for all \(g \in G\).


That is, we just need to check that each \(gNg^{-1}\) is a subset of \(N\).

Proof
Given that \(gNg^{-1} \subseteq N\). We want to show that this implies that \(N \subseteq gNg^{-1}\) which means that \(gNg^{-1} = N\) which is what is required by the definition.

Let \(g \in G\) and \(n \in N\). Observe that

$$ \begin{align*} n &= (gg^{-1})n(gg^{-1}) \\ &= g(g^{-1}ng)g^{-1} \\ &= g(x)g^{-1} \end{align*} $$

where \(x = g^{-1}ng\). If we show that \(gNg^{-1} \subseteq N\), then we need to show that for any \(n \in N\), this \(n\) can be written as \(n = g^{-1}ng = x\) which means that we want to show that \(x \in N\). But by the hypothesis

$$ \begin{align*} x &= g^{-1}ng \\ &= g^{-1}n(g^{-1})^{-1} \\ &= ana^{-1} \quad \text{ (let $a = g^{-1}$ where $g^{-1} \in G$)} \end{align*} $$

Since \(g^{-1}\) is just an arbitrary element in \(G\), then \(ana^{-1} \in N\) which is what we wanted to show. \(\ \blacksquare\)


Examples

Take \(D_3 = \{e, r, r^2, j, rj, r^2j\}\) where \(r^3 = e = j^2\) and \(jr = r^{-1}j\). This rule also implies \(jr^k = r^{-k}j\). Now suppose we have the subgroup generated by \(j\), \(H = \langle j \rangle = \{e, j\}\). Is this a normal subgroup?

For every \(g \in G\), we want to show that \(gHg^{-1} \subseteq H\) so

$$ \begin{align*} eHe^{-1} &= H. \\ rHr^{-1} &= \{rer^{-1}, rjr^{-1}\} = \{e, r^2j\} \neq H. \end{align*} $$

So we stop here. This is not a normal subgroup. Does \(D_3\) have any normal subgroups? The trivial subgroups are normal subgroups. What about any non-trivial subgroups? It turns out that \(N = \langle r \rangle = \{e,r,r^2\}\) is normal in \(D_3\). We have to check all 6 cases

$$ \begin{align*} eNe^{-1} &= N \\ rNr^{-1} &= N \\ r^2Nr^{-2} &= N \\ jNr^{-1} &= N \\ rjN(rj)^{-1} &= N \\ r^2jN(r^2j)^{-1} &= N. \end{align*} $$

FACT: If \(G\) has some generating set like \(G = \langle g_1, ..., g_r \rangle\), then to check if \(H\) is normal, we just need to show that \(g_kHg_k^{-1}\) for each \(g_i\) in the generating set \(g_1, ... g_k\). So for \(D_3\), we know it is generated by \(\langle r, j \rangle\), so we can check those two.


Kernels are Normal Subgroups

Back to homomorphisms. It turns out that kernels of homomorphisms are normal subgroups.

Proposition
If \(\varphi \ : \ G \rightarrow H\) is a homomorphism, then \(K = ker \varphi = \{g \in G \ | \ \varphi(g) = e\} \) is a normal subgroup.


Proof
We know that \(k\) is a subgroup of \(G\). We need to show that given \(g \in G\) and \(x \in K\) that \(gxg^{-1} \in K\). By definition, to check if any element is in the kernel, we need to apply the homomorphism and see if the result is the identity element. So

$$ \begin{align*} \varphi(gNg^{-1}) &= \varphi(g)\varphi(x)\varphi(g)^{-1} \\ &= \varphi(g)e\varphi(g)^{-1} \\ &= \varphi(g)\varphi(g)^{-1} \\ &= e. \end{align*} $$

From this, we see \(gKg^{-1} \subseteq K\) for all \(g \in G\). \(\blacksquare\)

Note also that all subgroups of abelian groups are normal.


Injectivity of Homomorphisms

Another useful fact:

Proposition
Let \(\varphi \ : \ G \rightarrow H\) be a homomorphism, then \(\varphi\) is injective if and only if \(K = ker(\varphi) = \{e\} \).


Proof
\(\Longrightarrow\): Suppose that \(\varphi\) is injective. This means that if \(\varphi(a) = \varphi(e) = e\), then \(a = e\). But this means that the only element where \(\varphi(a) = e\) is the identity element itself. [why must \(varphi(e)=e\)? I can’t remember, verify].

\(\Longleftarrow\):Now suppose that \(K = ker(\varphi) = \{e\}\). We want to show that \(\varphi\) is injective. This means that for any elements \(a, b \in G\), if \(\varphi(a) = \varphi(b)\), then we must have \(a = b\). Let \(\varphi(a) = \varphi(b) = h \in H\). We want to show that \(a = b\). Observe that

$$ \begin{align*} \varphi(a^{-1}b) &= \varphi(a)^{-1}\varphi(b) \\ &= h^{-1}h \\ &= e_H. \end{align*} $$

Since \(\varphi(a^{-1}b) = e_H\), then this implies that \(a^{-1}b \in K\). But \(K\) has only a single element which is the identity. Then \(a^{-1}b = e\) which means that \(a = b\) as desired. \(\ \blacksquare\)



References