Show that $A_n$ is a normal subgroup of $S_n$ by defining a homomorphism from $S_n$ to $Z_2$ and showing that $A_n$ is its kernel.

Question

I know that the kernel is always a normal subgroup, it proves $A_n$ is a normal subgroup of $S_n$. but how to find the homomorphism.

Answer 1

I would define the function $\phi:S_n\rightarrow\mathbb{Z}_2$ by $$\phi(\alpha)=\begin{cases}1, \text{ if } \alpha\in A_n;\\-1\text{ if } \alpha\in O_n\end{cases}$$ where $O_n$ denotes the set of odd permutations.

It can be verified that $\phi$ is a homomorphism since product of odd and even permutation is always odd, and product of both odd or both even permutation is always even.
And clearly, from how we define $\phi$, we can see that $\ker \phi=A_n$.