Is the map $f:S_n \to A_{n+2}$ a homomorphism where $f(s)=s$ when $s$ is even and $f(s)=s \circ (n+1,\ n+2)$ when $s$ is odd?

Question

Is the map $f:S_n \to A_{n+2}$ given by

$$f(s)= \begin{cases} s & s\ \text{is even}\\ s \circ (n+1,\ n+2) & s\ \text{is odd} \end{cases}$$

an injective homomorphism? I can show that if it is a homomorphism then it is injective but having difficulty in showing that $f$ is a homomorphism. Please help.

Answer 1

As disjoint cycles commute and $\;(n+1\;n+2)^2=1\;$ , for any two cycles $\;\sigma,\pi\in S_n\;$ we get

$$f(\sigma\pi):=\begin{cases}\sigma\pi&,\text{both cycles have same parity}\\{}\\\sigma\pi(n+1\;n+2)=\sigma(n+1\;n+2)\pi=f(\sigma)f(\pi)&,\text{otherwise}\end{cases}$$

where $\;\sigma\;$ odd (first) case and $\;\pi\;$ even, and the other way around in the second case.

Now generalize using that any permutation is the product of disjoint cycles.

Answer 2

A map $f : S_n \to A_{n+2}$ is a homomorphism if $f(\sigma\circ\tau) = f(\sigma)\circ f(\tau)$ for every $\sigma, \tau \in S_n$. As $f$ is defined piecewise, we split this verification into cases.

1. Both $\sigma$ and $\tau$ are even. Note that $\sigma\circ\tau$ is even so

$$f(\sigma\circ\tau) = \sigma\circ\tau = f(\sigma)\circ f(\tau).$$

2. Both $\sigma$ and $\tau$ are odd. Note that $\sigma\circ\tau$ is even so

\begin{align*} f(\sigma\circ\tau) &= \sigma\circ\tau\\ &= \sigma\circ\tau\circ(n+1, n+2)\circ (n+1, n+2)\\ &= \sigma\circ(n+1, n+2)\circ\tau\circ(n+1, n+2)\\ &= f(\sigma)\circ f(\tau). \end{align*}

3. $\sigma$ is even and $\tau$ is odd. Note that $\sigma\circ\tau$ is odd so

$$f(\sigma\circ\tau) = \sigma\circ\tau\circ(n+1, n+2) = f(\sigma)\circ f(\tau).$$

4. $\sigma$ is odd and $\tau$ is even. Note that $\sigma\circ\tau$ is odd so

$$f(\sigma\circ\tau) = \sigma\circ\tau\circ(n+1, n+2) = \sigma\circ(n+1, n+2)\circ\tau = f(\sigma)\circ f(\tau).$$

So, for any $\sigma, \tau \in S_n$, $f(\sigma\circ\tau) = f(\sigma)\circ f(\tau)$, so $f$ is a homomorphism.

Answer 3

One more way to think about it:

Conceive of your map $f$ as actually being from $S_n$ to $S_{n+2}$. The image lies in the subgroup $S_n\times S_2$ consisting of permutations that act independently on the first $n$ and the last $2$ indices. You can see your map as being the identity map to the first factor ($S_n$), and exactly the sign homomorphism to the second factor ($S_2$). Therefore it is a direct product of homomorphisms, so it is a homomorphism.

It happens (because of how you constructed it) that the image lies inside $A_{n+2}$, so it's a homomorphism of $S_n$ into $A_{n+2}$.