I have started reading about subgroup growth and, to my surprise, I haven't found a reference to whether direct products preserve subgroup growth.

Recall that, given a finitely generated group $G$, the function $s_n(G)$ is given by
$$s_n(G)=\#\{\text{subgroups of } G\text{ of index }\leq n\}.$$
We say that $G$ has *subgroup growth type* $f$ for some function $f$ if there are $a$, $b>0$ such that
\begin{align}
s_n(G)&\leq f(n)^a \quad & &\text{for } n \text{ large enough},\\
f(n)&\leq s_n(G)^b \quad & &\text{for infinitely many } n.
\end{align}

What I have found is the following: Let $G$ be a finitely generated group, let $N$ be a normal subgroup and let $Q=G/N$. Then Proposition 1.3.2 in *Subgroup growth* by A. Lubotzky and D. Segal states
\begin{align}
s_n(G)&\leq s_n(Q)s_n(N)n^{\text{rk}(Q)},\\
s_n(G)&\leq s_n(Q)s_n(N)c^n,\qquad \text{where}\ c=3^{d(Q)/3}.
\end{align}
Of course, these inequalities can be applied to a direct product $G\times G$ by taking $N=G\times 1$.

This seems to suggest that there should exist a group $G$ of intermediate growth and infinite rank such that $G\times G$ has strictly faster subgroup growth than $G$; more precisely, given any $a>0$, $$s_n(G\times G)>s_n(G)^a \quad \text{for infinitely many } n.$$ Does such an example exist?