$G$ 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?
Best Answer
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.