The proof of $S_n \cong A_n \rtimes \{e, (12) \}$

Question

$\blacksquare~$ Problem: Let $G = S_n, H = A_n$ and $K = \{ e, (12) \}$. Show that $S_n \cong A_n \rtimes K$.

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$\blacksquare~$ My Approach:

Let $G = S_{n}$. Where $S_{n}$ is the symmetic group of order $n$.

The two groups $H = A_{n}$ and $K = \{ e , ( 1 , 2 ) \}$ . Clearly, $A_{n}$ and $\{ e , ( 1 , 2 ) \}$ are subgroups of $S_{n}$. We need to show that, \begin{align*} \begin{split} & 1) H \unlhd G\\ & 2) H \cap K = \{ e \} \end{split} \end{align*} At first we'll show that, $ A_{n} \unlhd S_{n} $.

$\bullet~$$\textbf{Proof:}~$ According to the definition of normal subgroups, for $ \sigma $ $\in$ $ S_{n}$ , $ \gamma $ $ \in $ $ A_{n}$, we have $ \sigma \gamma {\sigma}^{-1} $ $ \in $ $ A_{n} $.

Now, $ \sigma $ and $ {\sigma}^{-1} $ can be even or odd. And $\gamma $ is even. If $ \sigma $ is even then $ {\sigma}^{-1} $ is also even. And if $ \sigma $ is odd then also $ {\sigma}^{-1} $ is odd.

We can easily see that, when $\sigma$ is even, then we are done! But when $\sigma$ is odd, we have $\sigma \gamma {\sigma}^{-1}$ $\in$ $A_{n}$, as $\sigma \gamma {\sigma}^{-1}$ will be even. Hence $A_{n} \unlhd S_{n}$.

And we trivialy have $H \cap K = \{ e \}$ as $A_{n} \cap \{ e , (12) \} = \{ e \}$

Again, we have \begin{align*} \begin{split} \lvert HK \rvert & = \frac{\lvert H \rvert \lvert K \rvert }{ \lvert H \cap K \rvert } \\ & = \frac{\lvert A_{n} \rvert \lvert \{ e , ( 1 , 2 ) \} \rvert }{\lvert \{e\} \rvert}\\ & = \dfrac{\dfrac{n!}{2} \cdot 2}{1}\\ \\ & = n! = \lvert G \rvert\\ \end{split} \end{align*} Hence , we can see that, $H$ , $K$ $\leq$ $G$ and $\lvert HK \rvert $ = $\lvert G \rvert$. But $HK \subseteq G$.

Therefore , $G = HK$.

And from the definition of the semidirect product, we obtain that, $G \cong H \rtimes K$.

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Is the solution correct? Any new approach will be great!

Thanks in advance!

Answer 1

Your proof is fine.

A quicker way to do the first part might be to make use of the fact that conjugation of permutations preserves cyclic type (and hence parity), so that $\sigma\gamma\sigma^{-1}\in A_n$ for $\gamma\in A_n, \sigma\in S_n$.