Every group is solvable, fake proof.

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?

• Pffft, fake news! – Mathematician 42 Jan 31 '18 at 10:29
• The most serious mistake is that you appear to be assuming that all groups are finite which, for some reason that I do not understand, seems to be surprisingly common in this forum. – Derek Holt Jan 31 '18 at 11:49
• @DerekHolt: I suppose that nowadays a "tag" of (finite-groups) is regarded as taking the place of a hypothesis :-/ – Lee Mosher Jan 31 '18 at 14:08