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Let $p\geq 7$ be a prime and $m$ be a positive integer. Prove that group of order $6p^m$ is solvable.

Attempt:

By Sylow's theorems we have that $n_p \mid 6$ so $n_p\in \{1,2,3,6\}$ where $n_p$ is the number of Sylow $p$ groups. Also we have that $n_p \equiv 1 \pmod p$ so $n_p=1$ and is thus normal. So we investigate:

$$G\trianglerighteq H_p$$

Where $H_p$ denotes the Sylow $p$ group. We know that $|G/H_p|=6$ and there are two groups of order $6$: $\Bbb Z_6$ and $S_3$. Both are solvable. We also know that Sylow groups are solvable. Since $G/H_p$ is solvable and $H_p$ is solvable, $G$ is solvable.

Is this correct?

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  • $\begingroup$ Well, as you've noticed denoting as $\;S_7\;$ a Sylow subgroup is a very bad idea...so edit your question and change it! $\endgroup$ – DonAntonio Dec 7 '18 at 20:58
  • $\begingroup$ @DonAntonio there. $\endgroup$ – user608030 Dec 7 '18 at 21:16
  • $\begingroup$ you are using $7$ but you really mean $p$ $\endgroup$ – the_fox Dec 8 '18 at 3:52
  • $\begingroup$ @the_fox sorry, of course $\endgroup$ – user608030 Dec 8 '18 at 4:09
  • $\begingroup$ See also this question, with similar arguments. $\endgroup$ – Dietrich Burde Dec 8 '18 at 9:03
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By the way, there's no need to take $p\ge 7$. The result holds for all primes $p$. For $p=2,3$ this follows from Burnside's $p^aq^b$ theorem. For $p=5$ you additionally use the result that any group of order divisible by 2 but not 4 has an index 2 subgroup (of even permutations in the regular representation). And that proof also works for larger $p$ as well, and so is another solution to the original question

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  • $\begingroup$ See here for the result about groups of order $\equiv2\pmod4$. $\endgroup$ – Jyrki Lahtonen Dec 30 '18 at 14:23

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