If $G$ is the direct product of two cyclic $p$-groups, what is implied for the order of the automorphism group of $G$ Let $G$ be the direct product of two cyclic $p$-groups, then if $p \ge 3$ the automorphism group of $G$ is only divisble by primes $q \le p$, and if $p = 2$, then the only primes dividing the automorphism group are $2$ and $3$.
The question came up in my other post, and it was suggested in the comments that I should open a separate question. As noted in the comments, if the two factors are characteristic, then for all prime divisors $q$ of the automorphism group we have $q \le p$ (even if $p = 2$). But I do not see  how to generalise this to non-characteristic factors in the above stated manner?
 A: $\renewcommand{\phi}{\varphi}$$\newcommand{\GL}{\mathrm{GL}}$$\newcommand{\Aut}{\mathrm{Aut}}$Consider the Frattini subgroup $\Phi(G)$ of $G$, so that $G / \Phi(G)$ is elementary abelian of order $p^{2}$, that is, a vector space of dimension $2$ over the field with $p$ elements. (I am assuming $G$ to be the product of two nontrivial cyclic subgroups.)
The Frattini subgroup is characteristic, so restriction yields a morphism
$$
\phi : \Aut(G) \to \Aut(G/\Phi(G)).
$$
Now it can be proved that the kernel of $\phi$ is a $p$-group.
So if $q \ne p$ is a prime dividing the order of $\Aut(G)$, it must divide the order of $\GL(2, p)$, which is $(p^{2} - 1) (p^{2} - p) = p  \cdot (p-1)^{2} (p + 1)$. So either $q < p$, or $q = p + 1$, and thus $p = 2$, $q = 3$.
Note that if $G = A \times B$, with $A, B$ cyclic and non-trivial, neither factor can be characteristic. In fact, suppose $A = \langle a \rangle$, $B = \langle b \rangle$, with orders respectively $p^{n} \ge p^{m}$. Then the automorphism
$$
a \mapsto a b, b \mapsto b
$$
maps $A$ to $\langle a b \rangle \ne A$, and the automorphism
$$
a \mapsto a , b \mapsto b a^{p^{n-m}}
$$
maps $B$ to $\langle b a^{p^{n-m}} \rangle \ne B$.
