In this note, I've read that $\mathbb S_n$ is a semidirect product of the alternating group $A_n$ by $\mathbb Z_2$. So I am trying to define a morphism $\rho: \mathbb Z_2 \to Aut(A_n)$ to show that $\mathbb S_n \cong A_n \rtimes Z_2$. I would appreciate suggestions on how could I define the morphism. Thanks in advance.
Abstract Algebra – S_n as Semidirect Product
abstract-algebragroup-theorysemidirect-product
Related Solutions
Let $K \lhd G$. Then, $h^{-1}(k^{-1}hk)$ lies in $H$ as $H$ is normal, and $(h^{-1}k^{-1}h)k$ lies in $K$ as $K$ is normal. As their intersection is $1$, we have that $h^{-1}k^{-1}hk=1$, i.e. $hk=kh$.
Now, If $H$, $K$ are abelian, and $K \lhd G$, then the above computation shows that all elements of $G$ commute, and so $G$ is abelian.
The converse is trivial (why?)
Note that we do not require the actual automorphism $\rho$ for any of the above computations.
I find it helpful to distinguish between the inner semidirect product and the outer semidirect product. (Similiar to the distinction between the inner direct sum and the outer direct sum of vector (sub)spaces.)
Given a group $G$ and two subgroups $H$ and $N$ of $G$, the group $G$ is called the inner semidirect product of $H$ and $N$ with $N$ normal, if we have $NH = G$ and $H \cap N = 1$, and $N$ is normal in $G$. We then write $$ G = N \rtimes H \,. $$ (This notion of an inner semidirect product does not depend on any homomorphism.)
Given on the other hand any two groups $N$ and $H$ and a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$, the corresponding outer semidirect product $N \rtimes_\theta H$ is defined as the set $N \times H$ together with the multiplication $$ (n_1, h_1) \cdot (n_2, h_2) := (n_1 \theta(h_1)(n_2), h_1 h_2) \,. $$ Note that this group $N \rtimes_\theta H$ very much depends on $\theta$.
Now what is the connection between these two?
If $G$ is a group and $H$ and $N$ are two subgroups of $G$ such that $G = N \rtimes H$, then $N$ is normal in $G$. The conjugation action of $H$ on $G$ does therefore restrict to an action of $H$ on $N$. From this we get a group homomorphism $$ \theta \colon H \to \mathrm{Aut}(N) \quad\text{given by}\quad \theta(h)(n) = hnh^{-1} \,. $$ It can then be checked that the map $$ N \rtimes_\theta H \to G = N \rtimes H, \quad (n,h) \mapsto n \cdot h $$ is a group isomorphism. So every inner semidirect product can be seen as an outer semidirect product via the conjugation action of $H$ on $N$.
If on the other hand $N$ and $H$ are two groups and $\theta \colon H \to \operatorname{Aut}(N)$ a group homomorphism, then we can regard both $N$ and $H$ as subgroups $\widehat{N}$ and $\widehat{H}$ of the outer semidirect product $N \rtimes_\theta H$ via the inclusions \begin{alignat*}{2} N &\to N \rtimes_\theta H \,, &\quad n &\mapsto (n,1) \,, \\ H &\to N \rtimes_\theta H \,, &\quad h &\mapsto (1,h) \,. \end{alignat*} It then follows for those subgroup $\widehat{N}$ and $\widehat{H}$ that both $\widehat{N} \cdot \widehat{H} = \widehat{N} \rtimes_\theta \widehat{H}$ and $\widehat{N} \cap \widehat{H} = 1$, and that the subgroup $\widehat{N}$ is normal in $N \rtimes_\theta H$. The group $N \rtimes_\theta H$ is therefore the inner semidirect product of its subgroups $\widehat{N}$ and $\widehat{H}$, with $\widehat{N}$ normal.
PS: So given any two groups $H$ and $N$ there may exist multiple outer semidirect products $N \rtimes_\theta H$ because we can choose different group homomorphisms $\theta \colon H \to \operatorname{Aut}(N)$. But if we are given an inner semidirect product $G = N \rtimes H$ (i.e. we are given the group $G$ and its subgroups $H$ and $N$ such that $G = N \rtimes H$) then we are also implicitely given a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$ with $G \cong N \rtimes_\theta H$: The homomorphism $\theta$ is hidden in the group structure of $G$ and can be retrieved via the conjugation action of $H$ on $N$.
Best Answer
More generally:
Theorem. If $G$ has a subgroup of index two, $H$ say, and there exists an element $g\not\in H$ of order two then $G$ splits as a semidirect product $G=H\rtimes\mathbb{Z}_2$.
This is because $\langle g\rangle\cap H=1$, and the other (internal) semidirect product conditions follow because $H$ has index two (so is normal, and so on).
So, in order to answer your question you simply need to find an element of order two in $S_n$ which is not contained in $A_n$.