# Group of order $6p^m$ is solvable for prime $p\geq 7$

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?

• Well, as you've noticed denoting as $\;S_7\;$ a Sylow subgroup is a very bad idea...so edit your question and change it! – DonAntonio Dec 7 '18 at 20:58
• @DonAntonio there. – user608030 Dec 7 '18 at 21:16
• you are using $7$ but you really mean $p$ – the_fox Dec 8 '18 at 3:52
• @the_fox sorry, of course – user608030 Dec 8 '18 at 4:09
• See also this question, with similar arguments. – Dietrich Burde Dec 8 '18 at 9:03

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