Suppose that $|G|=pm$ where $p$ is a prime and $p\not|m$.
Prove that $gcd(p-1,m)=gcd(p,m-1)=1$ iff G has a normal Sylow p-subgroup in $Z(G)$
I have a problem in both directions.
How can I solve this problem with Sylow theorem?
2026-03-25 11:16:21.1774437381
$gcd(p,m-1)=gcd(p-1,m)=1$ if and only if G has a normal Sylow p-subgroup in $Z(G)$
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