Given $|G|=pq$, $p \leq q$ and $p,q$ prime. If $Z(G)=1$ and all non-identity have order $p$ then the centralizer of every group element has order $p$.

Asked Oct 27 '17 at 19:52

Active Sep 06 '19 at 22:13

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Assertion from Dummit and Foote 3rd p136: Given a group $G$ of order $pq$, $p \leq q$ and $p,q$ are prime. If $Z(G)=1$ and all non-identity has order $p$ then the centralizer of every group element has order $p$.

I can't follow the argument: I understand that centralizer must have order $p$ or $q$ but why it cannot have $q$ elements here. It doesn't look very obvious to me.

edited Sep 06 '19 at 22:13

Shaun

44,997

asked Oct 27 '17 at 19:52

Daniel Li

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2

What is the order of a non-identity element in a (sub)group of order $q$? – Arturo Magidin Oct 27 '17 at 19:55
oh I see... thank you – Daniel Li Oct 27 '17 at 19:57
Aside: $G$ has trivial center here iff $G$ is non-abelian: https://math.stackexchange.com/questions/999260/verification-of-proof-that-a-nonabelian-group-g-of-order-pq-where-p-and-q-are-pr. – Dietrich Burde Oct 27 '17 at 20:18

Given $|G|=pq$, $p \leq q$ and $p,q$ prime. If $Z(G)=1$ and all non-identity have order $p$ then the centralizer of every group element has order $p$.

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