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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

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  • $\begingroup$ Idea: $o(x_1,x_2)=\mathrm{lcm}\left(o_{G_1}(x_1),o_{G_2}(x_2)\right)$ $\endgroup$ – Peter Košinár Dec 4 '14 at 12:26
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    $\begingroup$ This seems to be a duplicate of this question which, however, does not yet have an answer. $\endgroup$ – bof Dec 4 '14 at 12:32
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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.

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  • $\begingroup$ Please elaborate on how to derive the inequality $\endgroup$ – Souvik Dey Dec 4 '14 at 12:40
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    $\begingroup$ Say $n_i = n(G_i, d)$ for short. We want to know the number of elements on the form $(x, y)$ where $x$ and $y$ are either of order $d$ or the identity in their respective groups, except for the one pair where they're both the identity. Such a pair will have order $d$. There are $(n_1 + 1) \cdot (n_2 + 1) - 1$ such pairs, and that's what I've written down. $\endgroup$ – Arthur Dec 4 '14 at 12:44

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