# Exercise 6.3.14 from Scott, Group Theory

G is a group, p a prime number; if $|G|=\ p^{2}$ then if G is cyclic |Aut(G)|=$\ p(p-1)$, if G is not cyclic |Aut(G)|=$\ (p+1)p(p-1)^{2}$. The part about the cyclic case is easy, but the other?

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## 1 Answer

If $G$ fulfills $|G|=p^2$ and $p$ prime then $G$ is necessarily Abelian (This is a pretty important result for this exercise but I would guess that it is a theorem or lemma somewhere in your book). Let us now assume that $G$ is not cyclic so we have an elementary Abelian group. Then $G = \mathbb{F}_p^2$ (up to isomorphism) and $G$ is basically a 2-dimensional Vectorspace over the field with $p$ elements.

With these hints you should probably be able to complete the second part of the proof. If there are any further questions just ask.

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So Aut(G) is GL(2,p), and then |Aut(G)|=(p+1)p(p-1)^2, exact? –  WLOG Dec 20 '11 at 8:35
Correct. If this was a homework exercise or a question in an exam you should probably explain why every vectorspaceautomorphism is a groupautomorphism and vice versa. –  Sebastian Schoennenbeck Dec 20 '11 at 9:38