A group is solvable if it has only one $p$-Sylow subgroup for each $p$

Question

The problem is to prove that a group is solvable if it has just one $p$-Sylow subgroup for each prime $p$ dividing its order.

My solution:

If$|G|=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$ proof by induction on $n.$ $n=1, |G|=p_1^{\alpha_1}$ so we know that is solvable. If we assume that for $n-1$ it is true, then for $n$, $G$ has a subgroup of order $p_1^{\alpha_1}$ by Sylow's theorem, and it gonna be normal, because it is only $p_1$-Sylow subgroup. Let's call it $H_1$. By induction hypothesis $G/H_1$ is solvable, and by base of induction $H_1$ is solvable. We conclude $G$ is solvable.

Is this okay? This is my homework problem, so I want to be sure.

Answer 1

This follows from two theorems:

• Why is a unique Sylow p-subgroup normal?

• If all Sylow subgroups are normal then the group is solvable