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(Post has been edited )

So I am reading the proof of this statement:

Let G be a group of order $pq$ with $p,q$ primes and $p<q$.

If $p \nmid (q-1)$, then $G\simeq \mathbb{Z}_{pq}$

if $p \nmid (q-1)$ then thre are two isomorphism classes: $\mathbb{Z}_{pq}$ and a certain non-abelian group.

So, here is part of the proof (until the part that I have problems with):

Supposing that $p \mid (q-1)$, let $P \in Syl_p(G), Q \in Syl_q(G)$, since $\mid G:Q\mid = p$, which is th least prime number that divides $\mid G \mid$, then $Q \unlhd G$. Because $(\mid P \mid, \mid Q \mid)=1$, then $P \cap Q = \{e\}$, so $G=QP$.

Then $G$ is a semidirect product $Q \rtimes_{\theta} P$ for a certain morhpism $\theta : P \to Aut(Q)$.

Now they assert that $Aut(Q) \simeq \mathbb{Z}_{q-1}$.

Why is this? I am pretty sure this is not true for any group since complete groups are isomorphic to their automorphim group by definition.

Any help would be appreciated.

Bajo Fondo
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  • Note that you have the conditions the wrong way round in your statement. The automorphism group of the cyclic group of order $q$ has to contain an element of order $p$ for the nonabelian construction to work. The existence of a primitive root modulo $q$ means that the automorphism group of $\mathbb Z_q$ is cyclic, and there will be only one subgroup of order $p$ (in the case that $p|q-1$). – Mark Bennet Jun 09 '18 at 16:38
  • @Mark Bennet you are right, I am going to edit it. – Bajo Fondo Jun 09 '18 at 18:22
  • The other part of my comment is intended to be helpful, so I'm not deleting it. @DietrichBurde was linked a useful duplicate where you will find expanded exposition. – Mark Bennet Jun 09 '18 at 21:22

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