Let $G$ be a group such that $|G:Z(G)|=pq$ with $p

Question

I need help with this statement:

Let $G$ be a group such that $|G:Z(G)|=pq$ with $p<q$ primes. Prove that $q \equiv 1 \pmod p$.

I took a subgroup $H$ of $G/Z(G)$ of order $p$ and I tried to apply the result $$|N_G(H):H| \equiv |G:H|\pmod p$$ but I'm stuck.

Any help is apreciated.

You used the tag [tag:finite-groups] but you don't mention finite groups in the question . . . — Shaun, Apr 17 '23 at 22:28
You're welcome, @olasiu, but please [edit] the question to make it clear exactly where you are using finite groups. — Shaun, Apr 25 '23 at 22:03

score 2 · Accepted Answer · answered Apr 18 '23 at 15:19

2

By contradiction, suppose $q\not\equiv 1\pmod p$. Then $p\nmid q-1$. A group of order $pq$, with $p\nmid q-1$, is cyclic (in the link, the roles of $p$ and $q$ are swapped). But if $G/Z(G)$ is cyclic, then $G$ is abelian, and hence $|G/Z(G)|=1$. Contradiction.

answered Apr 18 '23 at 15:19

citadel

2,940

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Thank you so much! – alfita Apr 25 '23 at 21:55

Let $G$ be a group such that $|G:Z(G)|=pq$ with $p

1 Answers1