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Is a group with order $16*17$ solvable?

I know that from Burnside this is solvable since 2 and 17 are prime and 4 is greater than 0. However, I am not allowed to use it, so what should I do?

Thanks in advance

I noticed that there is a similar thread Group of order $8p$ is solvable, for any prime $p$

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Calico
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  • Have you tried to deduce anything whatsoever? What kinds of things do we normally do with arithmetic and the order of the group, or at least try to see if we can do? – anon Dec 13 '14 at 04:56
  • @ anon I am taking abstract, and due to time conflict, I missed a couple classes on this, so I had to ask some obvious questions. – Calico Dec 13 '14 at 05:00

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Number of Sylow-$17$ subgroup is congruent to $1$ mod $17$, and it divides $2^4$, hence it must be $1$; so Sylow-$17$ subgroup is unique, hence normal, call it $P$. Then $G/P$ is group of order $2^4$; since $p$-groups are solvable, it follows that both $P$ and $G/P$ are solvable, hence $G$ is solvable.

Groups
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  • Which theorem did you use to know that the number of Sylow-17 subgroup is congruent to 1 mod 17, and it divides 24? From my textbook, I know that if X is congruent to 1 mod 17, for X the number of congugate of a Sylow group. – Calico Dec 13 '14 at 05:15
  • Oh, never mind. I got it. Thanks very much. – Calico Dec 13 '14 at 05:16
  • @Calico: welcome! – Groups Dec 13 '14 at 05:17
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    Better form of a Sylow Theorem (as Originally Sylow states): If $P$ is a Sylow $p$-subgroup of $G$, then the number of Sylow $p$-subgroup divides $|G|/|P|$. In your question, $|G|=2^4.17$, and $|P|=17$. – Groups Dec 13 '14 at 05:38