TITLE: A problem in group theory_dsom
QUESTION [5 upvotes]: Let $H$ be a group of integers modp, under addition, where $p$ is a prime number. Suppose that $n$ is an integer satisfying $1 \leq n \leq p$, and let G be the group $ H \times H \times \cdots \times H$ (n factors). Show that $G$ has no automorphism of order $p^2$.

REPLY [4 votes]: $H$ is in fact a field, hence $G$ can be viewes as a vector space over $H$, and every automorphism of the group $G$ is also a linear automorphism of the vector space $G$. As such it can be described by an $n\times n$ matrix $A$. If $G$ has order $p^2$ then this matrix has an eigenvalue (in an algebraic closure of $H$) that is a primitive $p^2$th root of unity. The degree of its minimal polynomial is $\phi(p^2)=p\cdot(p-1)$ and this larger than the degree $\deg(A-\lambda)$ of the characteristic polynomial - unless $p=n=2$. But the automorphism group of $C_2\times C_2$ is well-known and equals $S_3$, which does not contain a $C_4$.