Is the symmetric group S3 a direct product of nontrivial groups? Is the symmetric group S3 a direct product of nontrivial groups? I was told by someone that it is the semidirect product of a nonabelian group and an abelian group, but I still dont know why. Can someone elaborate it?
Thank you!
 A: Suppose $S_3 \cong A \times B$ where neither $A$ nor $B$ is trivial. Then $6 = \lvert S_3 \rvert = \lvert A \rvert \lvert B \rvert$ and neither $\lvert A \rvert$ nor $\lvert B \rvert$ is $1$. Thus, WLOG, $\lvert A \rvert = 2$ and $\lvert B \rvert = 3$. This implies $A \cong \mathbb{Z}/2\mathbb{Z}$ and $B \cong \mathbb{Z}/3\mathbb{Z}$. In particular, $A$ and $B$ are abelian, so $S_3 \cong A \times B$ is also abelian. This contradicts the fact that $S_3$ is not abelian.
A: The symmetric group $S_3$ has 6 elements: $S_3=\{e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)\}$.
Since $(1 2)(1 3)=(1 2 3)$ and $(1 3)(1 2)=(1 3 2)$, this group $S_3$ is not abelian. Suppose $S_3$ can be written as a direct product of non trivial groups A and B.
Then order of A and order of B are not equal to one.
So the possible orders of the groups A and B are 2, 3.
Without loss of generality, take order of A =2 and order of B =3. Thus $A \cong \mathbb{Z}_2$ and $B \cong \mathbb{Z}_3$. Therefore both A and B are abelian and so $S_3$ is abelian (since the direct product of two abelian groups is also abelian). This is a contradiction as we have shown that $S_3$ is not abelian.
