let $(G,*)$ a group order $p^2q^2$ such that $q\nmid p^2 -1 $ y $p\nmid q^2 -1$ then $G$ is abelian. for Sylow theorem $n_p\equiv 1\mod (p)$ then $n_p = 1, q, q^2 $ but $n_p\neq p^2$ the same form $n_q \neq q^2$ but what next??
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$(q-1)\mid(q^2-1)$, so $p\nmid(q-1)$ – Jyrki Lahtonen Sep 03 '12 at 17:59
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That should be $n_p = 1,q,q^2$ (not $1,p,p^2$). – Ted Sep 03 '12 at 18:04
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2Possible duplicate of A group of order $p^2q$ will be abelian – Dietrich Burde Oct 04 '16 at 21:01
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You can show that your group $G$ is as $H\times K$ wherein $H$ and $K$ are normal sylow $p$-subgroup and normal sylow $q$-subgroup respectively. Now, since $H$ and $K$ are abelian groups, so is $G$ itself.Use what Jyrki noted above to show that you have only one sylow $p$-subgroup and one sylow $q$-subgroup.
Mikasa
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