# The Symmetric Group is a Semi-Direct Product of the Alternating Group and a Subgroup $\langle(1,2) \rangle$ ## Problem 465

Prove that the symmetric group $S_n$, $n\geq 3$ is a semi-direct product of the alternating group $A_n$ and the subgroup $\langle(1,2) \rangle$ generated by the element $(1,2)$. Add to solve later

## Definition (Semi-Direct Product).

### Internal Semi-Direct-Product

Recall that a group $G$ is said to be an (internal) semi-direct product of subgroups $H$ and $K$ if the following conditions hold.

1. $H$ is a normal subgroup of $G$.
2. $H\cap K=\{e\}$, where $e$ is the identity element in $G$.
3. $G=HK$.

In this case, we denote the group by $G=H\rtimes K$.

### External Semi-Direct Product

If $G$ is an internal semi-direct product of $H$ and $K$, it is an external semi-direct product defined by the homomorphism $\phi:K \to \Aut(H)$ given by mapping $k\in K$ to the automorphism of left conjugation by $k$ on $H$.
That is, $G \cong H \rtimes_{\phi} K$.

## Proof.

Recall that each element of the symmetric group $S_n$ can be written as a product of transpositions (permutations which exchanges only two elements).
This defines a group homomorphism $\operatorname{sgn}:S_n\to \{\pm1\}$ that maps each element of $S_n$ that is a product of even number of transpositions to $1$, and maps each element of $S_n$ that is a product of odd number of transpositions to $-1$.

The alternating group $A_n$ is defined to be the kernel of the homomorphism $\operatorname{sgn}:S_n \to \{\pm1\}$:
$A_n:=\ker(\operatorname{sgn}).$

As it is the kernel, the alternating group $A_n$ is a normal subgroup of $S_n$.
Also by first isomorphism theorem, we have
$S_n/A_n\cong \{\pm1\},$ and it yields that
$|A_n|=\frac{|S_n|}{|\{\pm1\}|}=\frac{n!}{2}.$

Since $\operatorname{sgn}\left(\,(1,2) \,\right)=-1$, the intersection of $A_n$ and $\langle(1,2)\rangle$ is trivial:
$A_n \cap \langle(1,2) \rangle=\{e\}.$

Let $H=A_n$ and $K=\langle(1,2) \rangle$.
Then we have
\begin{align*}
|HK|=\frac{|H|\cdot |K|}{|H\cap K|}=|H|\cdot | K|=\frac{n!}{2}\cdot 2=n!.
\end{align*}
Since $HK < S_n$ and both groups have order $n!$, we have $S_n=HK$.

In summary we have observed that $H=A_n$ and $K=\langle(1,2) \rangle$ satisfies the conditions for a semi-direct product of $G=S_n$.
Hence
$S_n=A_n\rtimes \langle(1,2) \rangle.$

As an external semi-direct product, it is given by
$S_n \cong A_n\rtimes_{\phi} \langle(1,2) \rangle,$ where $\phi: \langle(1,2) \rangle \to \Aut(A_n)$ is given by
$\phi\left(\, (1,2) \,\right)(x)=(1,2)x(1,2)^{-1}.$ Add to solve later

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