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I need to prove that a group $G$ with $|G| = pq$, where $p$, $q$ are primes, cannot be simple. I have already reduced this problem to showing that a unique Sylow $p$-subgroup is normal. The answers i have found so far are something along the line of "the Sylow $p$-subgroup is normal because all $p$-Sylow subgroups are conjugate to each other" which means diddly-squat to me. I need help understanding that last part.

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  • $\begingroup$ I think it's because if it's the only subgroup of that order, then it's characteristic (every automorphism fixes that group), and all characteristic subgroups are normal :) $\endgroup$ – Foobanana Nov 2 '17 at 16:14
  • $\begingroup$ $N$ is normal in $G$ if and only if $gNg^{-1}=N$ for all $g \in G$. This is exactly saying that $N$ always conjugates to itself. If there is only one such group and conjugation preserves order, guess what? $\endgroup$ – Randall Nov 2 '17 at 16:14
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    $\begingroup$ You don't need the fact that all Sylow $p$-subgroups are conjugate. It's simpler than that. If $H$ is the unique Sylow $p$-subgroup, then it is the only subgroup of that order. Any automorphism (including conjugation) of $G$ must map $H$ to another subgroup of the same order. Since $H$ is the only such subgroup, $H$ must be mapped to itself. In particular, the conjugate of $H$ is itself, so $H$ is normal. $\endgroup$ – Bungo Nov 2 '17 at 16:15
  • $\begingroup$ I think @Bungo put it better than I did :) $\endgroup$ – Foobanana Nov 2 '17 at 16:16
  • $\begingroup$ @Foobanana Your remark is quite correct, I just put a bit more detail :-) $\endgroup$ – Bungo Nov 2 '17 at 16:17
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Suppose that $H$ is the unique $p$-Sylow subgroup, for any $g\in G, gHg^{-1}$ is also a $p$-Sylow subgroup since it has the same cardinal than $H$, since $H$ is the unique $p$-Sylow subgroup, $gHg^{-1}=H$ so for every $h\in H, ghg^{-1}\in H$ and $H$ is normal.

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