Let $H$ be the group of integers mod p, under addition, where $p$ is a prime number. Suppose that n is an integer satisfying $1 ≤ n ≤ p$, and let G be the group $H × H × · · · ×H $($n$ factors). Show that $G$ has no automorphism of order $p^2$.
This is a problem of Berkeley spring $83$ prelim exam. The same question had been asked here, but there were no answers. Therefore I am asking it again. Can anyone please help me out ?
I have understand the comments of the old problem. Sylow-$p$ subgroup of $GL_n(\mathbb{F}_p)$ will have order $p^\frac{n(n-1)}{2}$. I am unable to proceed further from here.
Let $F_p$ be the field with $p$ elements, let $n$ be an integer such that $1\leq n\leq p$. Your group $G$ is the abelian group underlying the $F_p$ vector space $F_p^n$, and an automorphism of that abelian group is the same as an automorphism of that vector space. This means that your problem is the same as:
Now let $f:F_p^n\to F_p^n$ be an automorphism of order $p^2$. Then $f^{p^2}=I$, so the polynomial $X^{p^2}-1$ vanishes on $f$. That polynomial factors as $(X-1)^{p^2}$, and this tells us that the linear map $f-1$ is nilpotent. As the dimension of $F_p^n$ is $n$, of course, the nilpotency index is at most $n$, and therefore $(f-1)^n=0$. In particular, we have that $f^p-1=(f-1)^p=0$, and the order of $f$ is at most $p$: this is a contradiction.