Exercise 3.5.12 (Subgroup of $A_n$ isomorphic to $S_{n-2}$)

Proof. We identify $S_{n-2}$ with the subgroup of the permutations of $S_n$ which fix $n-1$ and $n$.

Consider the map

If $\alpha$ is even, then $\phi (\alpha )=\alpha$ is even, and if $\alpha$ is odd, $\phi (\alpha )=\alpha \cdot (n-1n)$ is also even. In both cases, $\phi (\alpha )\in A_n$.

Since $\alpha \in S_{n-2}$ acts on $[[1,n-2]]$ and $(n-1n)$ on the complementary set $[[n-1,n]]$, $\alpha$ commute with $(n-1n)$:

Surprisingly, $\phi$ is a homomorphism: Let $\alpha ,\beta \in S_{n-2}$.

• If $\alpha$ and $\beta$ are even, then $\mathit{𝛼𝛽}$ is even, so

  $$\phi (\alpha )\phi (\beta )=\mathit{𝛼𝛽}=\phi (\mathit{𝛼𝛽}).$$

• If $\alpha$ is odd and $\beta$ is even, then $\mathit{𝛼𝛽}$ is odd, so

  $$\phi (\alpha )\phi (\beta )=\alpha (n-1n)\beta =\mathit{𝛼𝛽}(n-1n)=\phi (\mathit{𝛼𝛽}).$$

• If $\alpha$ is even and $\beta$ is odd, then $\mathit{𝛼𝛽}$ is odd, so

  $$\phi (\alpha )\phi (\beta )=\mathit{𝛼𝛽}(n-1n)=\phi (\mathit{𝛼𝛽}).$$

• If $\alpha$ and $\beta$ are odd, then $\mathit{𝛼𝛽}$ is even, so

  $$\phi (\alpha )\phi (\beta )=\alpha (n-1n)\beta (n-1n)=\mathit{𝛼𝛽}(n-1n)(n-1n)=\mathit{𝛼𝛽}=\phi (\mathit{𝛼𝛽}).$$

This shows that $\phi$ is a homomorphism.

Moreover $\phi$ is injective: If $\alpha \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\phi (\alpha )=()=1_{[[1,n]]}$. Therefore $\alpha \in A_n$, otherwise $\alpha (n-1)=n$. Thus $\phi (\alpha )=\alpha$, so $\alpha =()$. Thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ is trivial, so $\phi$ is injective.

Hence the image $\phi (S_{n-2})$ of $\phi$ is a subgroup of $A_n$ isomorphic to $S_{n-2}$. □