Proving that a group of order $99$ is abelian

$G$ is a group. $|G|$ = 99. I'm to show that $G$ is abelian.

$G$ has 2 normal Sylow-subgroups, $S_3$ and $S_{11}$.

Since the orders of $S_3$ and $S_{11}$ are primes, they are both cyclic and abelian.

Since the orders of $S_3$ and $S_{11}$ are co-primes, they intersect trivially and the direct product $S_3 \times S_{11}$ is abelian and cyclic.

Since they intersect trivially one can find an injective homomorphism $\phi: S_3\times S_{11} \to G$.

Here's the thing I don't get: how do I show that it's surjective?

• An injective map between two finite sets of the same order must be surjective. – Derek Holt Jun 10 '17 at 11:26
• But I thought $S_3\times S_{11}$ was of order 33? – Mathaniel Jun 10 '17 at 11:30
• Here $|G| = 3^2 \cdot 11$, so the Sylow $3$-subgroups of $G$ have order $3^2 = 9$. – Travis Willse Jun 10 '17 at 11:32
• @Travis so if $|G|$ had been 297 (=$3^3\times 11$) instead, $S_3 \times S_{11}$ would still have been isomorphic to $G$? – Mathaniel Jun 10 '17 at 11:36
• Yes, and for the same reason. But NB for $|G| = 297$ we cannot conclude that $G$ is abelian, as there are two nonabelian groups $H$ of order $27$, and for these $H \times S_{11}$ is nonabelian. – Travis Willse Jun 10 '17 at 12:49

The point is that $S_{11}$ must be normal. Otherwise by Sylow's third theorem, there'll be at least twelve conjugates of $S_{11}$ which is too many.
Therefore $G$ is a semidirect product of $S_3$ and $S_{11}$. The action of $S_3$ on $S_{11}$ is trivial, as $S_{11}$ has not automorphism of order $3$, so the semidirect product is direct.