# Group Homomorphisms

Recall from the Group Isomorphisms page that if $(G_1, *_1)$ and $(G_2, *_2)$ are two groups then an isomorphism between these groups is a bijective function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that:

(1)Furthermore, the groups $(G_1, *_1)$ and $(G_2, *_2)$ are said to be isomorphic if such an isomorphism exists.

We will now look at another type of function that can arise between two groups called a homomorphism. Homomorphisms are similar to isomorphisms except they're not required to be bijections.

Definition: Let $(G_1, *_1)$ and $(G_2, *_2)$ be two groups. A Homomorphism between these groups is a function $f : G_1 \to G_2$ such that for all $x, y \in G_1$ we have that $f(x *_1 y) = f(x) *_2 f(y)$. If such a function exists then $(G_1, *_1)$ and $(G_2, *_2)$ are said to be Homomorphic. |

Let's look at a simple example. 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:

(2)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 $\sigma, \tau \in S_n$:

(3)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.