# Automorphisms of $\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p$

Consider the abelian group $$G = \underbrace{\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p}_{n},$$ where $p$ is prime and $1\le n \le p$. I want to show that $G$ has no automorphism of order $p^2$.

I observe that group automorphism of $G$ is the same as linear isomorphism of $G$ as a $\mathbb{Z}/p$-vector space. So the question is the same as to prove that there is no $n\times n$ invertible matrix in $\mathbb{Z}/p$ that has order $p^2$.

My attempt is to calculate the order of $GL(n,\mathbb{Z}/p)$ and show that $p^2$ does not divide that order, but unfortunately this only works for $n\le 2$.

Any element of order $p^2$ lies in a $p$-Sylow subgroup. Since all $p$-Sylow subgroups are conjugated, it is enough to consider the canonical $p$-Sylow subgroup given by upper triangular matrices with diagonal $1,\dotsc,1$. But if $A$ is such a matrix, then $(A-1)^n=0$ and hence $(A-1)^p=0$, i.e. $A^p=1$.
• Your argument also shows that if $A^{p^k} = Id$, then $A$ has eigenvalue 1 with multiplicity $n$. Is there a direct way to see this? – mez Jul 26 '14 at 14:19
No-Sylow proof: Just conjugate over $\bar{\mathbb{F}}_q$ to reach the Jordan normal form of this element, then the remaining is the same as Martin Brandenburg's answer.
A variant on user148212' s argument . Let $g$ be a $p$-element in $G$, that is, $g^{p^{k}} = 1$ for some $p$. Since we are in characteristic $p$, we have $(g - 1)^{p^{k}} = 0$, that is, $g-1$ is nilpotent. Then proceed as in Martin Brandenburg's answer, $(g-1)^{p} = 0$, so that reversing the argument $g^{p} = 1$.