$S_3$ is indecomposable but not simple

Question

My argument goes like this:

$S_3$ is not simple as $A_3$ is a subgroup of index $2$. On the other hand, if $S_3$ is decomposable, then $S_3 = A \times B$, where $|A| = 2$ and $|B| = 3$. But in this case $A = \mathbb{Z}_2$ and $B = \mathbb{Z}_3$, in which case its product is the cyclic group of order $6$; contradiction.

Is this right?

Answer 1

Yes your argument is right! As there are only two groups of order 6 (up to isomorphism) one is $\Bbb{Z}_6$ which is cyclic and other is $S_3$ which is non-abelian and can not be written as direct product of two of it's proper subgroup otherwise $S_3$ will be isomorphic to $\Bbb{Z}_6$ !