Group Homomorphisms

# Group Homomorphisms

 Definition: Let $(G, \cdot)$ and $(H, *)$ be two groups. A Homomorphism between these groups is a function $f : G \to H$ such that for all $x, y \in G$ we have that $f(x \cdot y) = f(x) * f(y)$.
 Definition: Let $(G, \cdot)$ and $(H, *)$ be two groups. Then: 1) A Monomorphism from $G$ to $H$ is a homomorphism $f : G \to H$ that is injective. 2) An Epimorphism from $G$ to $H$ is a homomorphism $f : G \to H$ that is surjective. 3) An Isomorphism from $G$ to $H$ is a homomorphism $f : G \to H$ that is bijective.

We will now look at some examples of group homomorphisms.

## Example 1

Let $(G, \cdot)$ and $(H, *)$ be groups. The direct product of $G$ and $H$ is the group consisting of the set $G \times H$ with the operation defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by:

(1)
\begin{align} \quad (g_1, h_1)(g_2, h_2) = (g_1 \cdot g_2, h_1 * h_2) \end{align}

Let $\pi_1 : G \times H \to G$ and $\pi_2 : G \times H \to H$ be defined by:

(2)
\begin{align} \quad \pi_1(g, h) &= g \\ \quad \pi_2(g, h) &= h \end{align}

Then $\pi_1$ is a homomorphism from $G \times H$ to $G$, and $\pi_2$ is a homomorphism from $G \times H$ to $H$.

## Example 2

For $n \geq 3$ consider the symmetric group whose operation is function composition $(S_n, \circ)$, the the group $(\mathbb{Z}_2, +)$. Define a function $f : S_n \to \mathbb{Z}_2$ for all $\sigma \in S_n$ by:

(3)
\begin{align} \quad f(\sigma) = \left\{\begin{matrix} 0 & \mathrm{if} \: \sigma \mathrm{\: is \: an \: even \: permutation} \\\ 1 & \mathrm{if} \: \sigma \mathrm{\: is \: an \: odd \: permutation} \end{matrix}\right. \end{align}

We will show that $f$ is a group homomorphism between $(S_n, \circ)$ and $(\mathbb{Z}_2, +)$. Let $\sigma, \tau \in S_n$. Then either: both $\sigma$ and $\tau$ are even permutations; both $\sigma$ and $\tau$ are odd permutations; or, one of $\sigma$ or $\tau$ is an even permutation and the other is an odd permutation.

Case 1: Suppose that $\sigma$ and $\tau$ are even permutations. Then $\sigma$ and $\tau$ can both be written as a product of an even number of transpositions, so $f(\sigma) + f(\tau) = 0 + 0 = 0$. But also $\sigma \circ \tau$ can be written as a product of an even number of transpositions. Therefore $f(\sigma \circ \tau) = 0$.

Case 2: Suppose that $\sigma$ and $\tau$ are odd permutations. Then $\sigma$ and $\tau$ can both be written as a product of an odd number of transpositions, so $f(\sigma) + f(\tau) = 1 + 1 = 0 \pmod 2$. But also $\sigma \circ \tau$ can be written as a product of an even number of transpositions. Therefore $f(\sigma \circ \tau) = 0$.

Case 3: Without loss of generality assume that $\sigma$ is an even permutation and $\tau$ is an odd permutation. Then $f(\sigma) = 0$ and $f(\tau) = 1$. So $f(\sigma) + f(\tau) = 0 + 1 = 1$. But also $\sigma \circ \tau$ can be written as a product of an odd number of transpositions. Therefore $f(\sigma \circ \tau) = 1$.

In all three cases we see that for all $\sigma, \tau \in S_n$:

(4)
\begin{align} \quad f(\sigma \circ \tau) = f(\sigma) + f(\tau) \end{align}

So $f$ is a homomorphism between $(S_n, \circ)$ and $(\mathbb{Z}_2, +)$. Of course $f$ is not an isomorphism though. This is because $f$ is not injective since for $n \geq 3$, $\mid S_n \mid \geq 6$, and $\mid \mathbb{Z}_2 \mid = 2$, and a function between finite groups of different sizes cannot be bijective.