Every group is solvable, fake proof.

Question

We use induction on $n$, the number of distinct prime factors of the order of $G$.

If $n=1$ then $G$ is a $p$-group which is solvable

Now assume this for some $k$. Let $G$ be a group of order with $k+1$ distinct prime factors. Then $G$ has a sylow p-subgroup, $H$. Then $\dfrac{G}{H}$ has order with $k$ distinct prime factors, so its solvable by our assumption. But $H$ is also solvable hence $G$ is solvable wich completes the induction step.

Thus every group is solvable. Where did I make a mistake?

Answer 1

A Sylow subgroup of a group is normal if and only if it’s the only one of its kind.