Group of order $3^k5^l$ is solvable

Asked Mar 15 '17 at 21:09

Active Sep 14 '19 at 01:48

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So I'm not sure I'm proving this right and would appreciate a correction if needed.

Let $G$ be a group of order $3^k5^l$, where $k,l\in\mathbb N$ and $k\le 3$. Prove that $G$ is solvable.

Proof:

Suppose $k=0$ or $l=0$ then the group is a $p$-group and therefore solvable. So $1\le k \le 3$. Assume that $l \ge 1$. Let $P \le G $ a $5$-sylow group if $P\triangleleft G$ then $P$ and $G/P$ are solvable and then $G$ is solvable. Now assume that $k=1$ then $n_5(G) | 3$ and $n_5(G)\equiv 1 (mod 5)$ so $n_5(G)=1$ so we have a normal $5$-sylow group. with $k=2$ $n_5(G) | 9$ and again $n_5(G)=1$, with $k=3$ $n_5(G) | 27$ and yet again $n_5(G)=1$. and we have for every $l\ge 1$ and $1\le k\le 3$ a normal sylow-$5$ group.

This seems "too easy" and I feel I made a mistake somewhere though I don't know where. Thank you for your help!

edited Sep 14 '19 at 01:48

Shaun

47,747

asked Mar 15 '17 at 21:09

Kim Seel

Compare with this question. – Dietrich Burde Mar 15 '17 at 21:12
2

Looks fine to me. Fun fact, the Burnside theorem states that any finite group of order $p^aq^b$ is solvable. But the proof uses representation theory and is very different from the above. – Mathematician 42 Mar 15 '17 at 21:14

Group of order $3^k5^l$ is solvable

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