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?

  • 1
    $\begingroup$ Pffft, fake news! $\endgroup$ – Mathematician 42 Jan 31 '18 at 10:29
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 31 '18 at 11:49
  • $\begingroup$ @DerekHolt: I suppose that nowadays a "tag" of (finite-groups) is regarded as taking the place of a hypothesis :-/ $\endgroup$ – Lee Mosher Jan 31 '18 at 14:08

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

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    $\begingroup$ Because all sylow subgroups are conjugate, I remember now :( $\endgroup$ – Noel Lundström Jan 31 '18 at 10:34

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