Proving a group of order $35^3$ is solvable

Question

Is my thinking correct when asked to show that a group $G$ of order $35^3$ is solvable, I first show that by the sylow theorems there exists a sylow $p$-subgroup of order $5^3$ and another unique sylow $p$-subgroup of order $7^3$. Then since these two unique sylow $p$-subgroups compose the group and are solvable then the entire group is solvable.

What are the techniques of showing a group is solvable with the sylow theorems?

I'm having trouble proving this one and another where the group is of order $80$.

Thank you for the help.

Answer 1

All of the Sylow subgroups of a group are unique if and only if it is the direct product of those subgroups.

But Sylow subgroups are $p $-groups; which are solvable.

Finally the direct product of solvable groups is again solvable.

Note that for a group of order $80$ it suffices to get one normal Sylow, since then it and the quotient group by it are $p $-groups.

Answer 2

If $n_5=1$, then let $P$ be the unique $5$-Sylow subgroup. Both $P$ and $G/P$ are $p$-groups, so $G$ is solvable.

If $n_5=16$, then each $5$-Sylow subgroup of $G$ has trivial intersection with each other $5$-Sylow subgroup so there are $4 \cdot 16 = 64$ elements of order $5$ in $G$. Since a $2$-Sylow subgroup must have $15$ non-identity elements, $G$ must have a unique $2$-Sylow subgroup, $Q$. Both $Q$ and $G/Q$ are $p$-groups, so $G$ is solvable.