This is Exercise 3.3.2 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.
The Details:
Robinson's book defines indecomposable groups via $\Omega$-subgroups by setting $\Omega=\varnothing$ in the definition of $\Omega$-indecomposable, so it is rather complicated; however, it is equivalent to Roman's definition on page 75 of "Fundamentals of Group Theory: An Advanced Approach":
A nontrivial group $G$ is said to be indecomposable if $G$ cannot be written as an internal direct product of two proper subgroups, that is, $$G=H ⨝ K\implies ((H=G)\lor(K=G)).$$
The Question:
Prove that $S_n$ is indecomposable.
Thoughts:
My first thoughts were as follows . . .
Suppose $S_n=H ⨝ K$ for proper $H,K\le S_n$ such that $H\neq S_n$. My aim is then to show that $K=S_n$.
That $K\subseteq S_n$ is clear from $K\le S_n$.
Let $\sigma\in S_n$. I require $\sigma\in K$ in order for $S_n\subseteq K$. Here, however, I am stuck; I mean: I could try & see where the statement that $H\neq S_n$ comes into play but I have no ideas there.
Thus I decided to try induction on $n$. (This might work since, by convention, $n$ is assumed to be finite.)
The group $S_1$ is degenerate here.
However, $S_2\cong \Bbb Z_2$ is clearly indecomposable. (It's overkill but this follows from the Fundamental Theorem of Finitely Generated Abelian Groups.)
Suppose $S_r$ is indecomposable for some $r\in \Bbb N\setminus\{1\}$.
Consider when $n=r+1$. We have some $T\le S_{r+1}$ such that $T\cong S_r$. Where do I go from here?
My intuition is more in line with my first thoughts, for I can sense in some way that $H\neq S_n$ would force $H\cong 1$ and $K=S_n$. But how do I show this?
I am surprised that this property of $S_n$ is not easy to find online. It has me wondering whether it's even true or, worse, whether it's somewhat trivial.
Please help :)
For $n=1$, it is trivial.
For $n=2$, since $S_n$ is a cyclic group of prime order, the result is also obvious.
Let $n\geq 3$. Suppose that there exists proper subgroups $H,K$ such that $S_n$ is a internal direct product of $H$ and $K$. Then note that $H,K$ are proper normal subgroups of $S_n$ and $|S_n|=|H||K|$.
If $n\neq 4$, note that the only proper normal subgroup of $S_n$ is $A_n$. Since $S_n$ does not have a normal subgroup of order $2$, this is a contradiction.
Consider $n=4$. The only proper normal subgroups of $S_n$ are $A_n$ and the Klein $4$-group $V$. Since $S_n$ does not have a normal subgroup of order $2$ or $6$, this is also a contradiction.