Question. All groups of order 175 are abelian?
I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.
I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| \cdot |K|$.
Further I know that G happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.
Could somebody please explain:
Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?
Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivialy, or is it because they are both abelian too?
Anything else I should know?