Show that $A_n$ is the kernel of a group homomorphism of $S_n \rightarrow \{−1,1\}$.

Question

I know that the kernel is always normal but how to prove or show that $A_n$ is the kernel?

Answer 1

Try building an explicit homomorphism

$$\phi : S_n \longrightarrow \{-1, 1\} $$

Let us show that $\phi(\pi_i) = sgn(\pi_i)$ is an homomorphism as asked.

Remember that any permutation $\pi_i$ can be decomposed in a number of transpositions. Although this decomposition is not unique, it can be shown that if some permutation $\pi_a$ has a decomposition with an even amount of transpositions, then all decompositions of $\pi_a$ must use an even number of transpositions and $\pi_a$ is said to be even. Analogously, if a permutation $\pi_b$ has a decomposition with an odd number of transpositions, then all decompositions of $\pi_b$ have an odd number of transpositions and $\pi_b$ is said to be odd.

When then define $sgn(\pi_i)$ as $1$ if $\pi_i$ is even and $-1$ if $\pi_i$ is odd.

I leave it to you as an exercise to show that $sgn(\pi_i \circ \pi_j) = sgn(\pi_i)\times sgn(\pi_j)$

Now all you have to do is show that this homomorphism has $A_n$ as kernel.