If $n(G,d)$ be the no. of elements of order $d$ in a group $G$ , then can we find $n(G_1 \times G_2,d)$ in terms
of $n(G_1,d)$ and $n(G_2,d)$ ? Since $o(x)=o(x,e_2)$ for any $x \in G_1$ and similar is true for $G_2$ I can
conclude $n(G_1 \times G_2,d) \ge n(G_1,d)+ n(G_2,d) $ , but cannot proceed further . Please help . Strict
lower and upper bounds for $n(G_1 \times G_2,d)$ will also be appreciated . Thanks in advance
It's not enough to know it for only $d$, we need at least some information on the elements of orders that divide $d$.
For instance we have $n(\Bbb Z_4, 4) = 2$ and $n(\Bbb Z_2, 4) = n(\Bbb Z_3, 4) = 0$ , but $n(\Bbb Z_4\times \Bbb Z_2, 4) = 4$ while $n(\Bbb Z_4\times \Bbb Z_3, 4) = 2$.
The element $(x, y) \in G_1\times G_2$ has order $d$ iff $\operatorname{lcm}(o(x), o(y)) = d$. Which of course will happen if both elements are of order $d$ or one of them is of order $d$ and the other is the identity. This gives the lower bound $$n(G_1\times G_2, d)\geq (n(G_1, d) + 1)\cdot (n(G_2, d) + 1) - 1$$I believe equality is guaranteed if $d$ is a prime.