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Respected All.

I was solving problems on external direct product of groups from the book Contemporary Abstract Algebra by Gallian, while the following thought came to mind.

The problem was " How many elements of order 2 are there in $\mathbb Z_2\times \mathbb Z_2 \times \mathbb Z_2$ ? I got the ans as 7.

From here I thought then what will happen for $\mathbb Z_3\times \mathbb Z_3\times \mathbb Z_3$ And in general $\mathbb Z_p\times \mathbb Z_p\times \mathbb Z_p$ if $p$ is any prime ?

Solving the last one was easy enough as $p^3-1$. Which helped me to get answer for $\mathbb Z_3\times \mathbb Z_3\times \mathbb Z_3$ directly.

From here I moved on to $\mathbb Z_p\times \cdots \times \mathbb Z_p$ ( n copies ): I guessed the answer should be $p^n-1$.

But I am unable to write it rigourously.

To use induction, we are assuming that at $n=1, 2, \cdots, m$ the result is true. Viz $$\eta_p(\mathbb Z_p\times \cdots \times \mathbb Z_p)=p^m-1$$ where $\eta_d(G)$ denotes the total number of elements of order d in the group $G$.

Now we have to establish the result for $n=m+1$ th level. Here I am asking help from you.

Suppose that we have to find out $\eta_d(G_1\times G_2)$ for the groups $G_1, G_2$ where it is given already that $\eta_d(G_1)=d_1, \eta_d(G_2)=d_2$. Can we get $\eta_d(G_1\times G_2)$ in terms of $d_1, d_2$ ?

If yes, then I will be able to complete my track.

Thanks in advance

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  • $\begingroup$ You are assuming that $d$ is a prime number, right? $\endgroup$ – bof Dec 4 '14 at 10:28
  • $\begingroup$ ummm....to be precise , no. I didn't think that. Just considered as any natural number/ positive integer. I want to know the result in general. $\endgroup$ – Anjan3 Dec 4 '14 at 11:11
  • $\begingroup$ Suppose $\eta_6(G_1)=\eta_6(G_2)=0$, what do you suppose that makes $\eta_6(G_1\times G_2)$? Consider some examples: $G_1$ and $G_2$ are groups of order $1$? $G_1=G_2=C_5$? $G_1=C_2$ & $G_2=C_3$? $\endgroup$ – bof Dec 4 '14 at 11:33
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    $\begingroup$ Anyway, your answer $p^n-1$ is for $p$ a prime number, right? Why don't you just prove that every element $g$ of $Z_p\times\cdots Z_p$ satisfies $g^p=1$, which means that each element has order $p$ or $1$ (since $p$ is prime), and only one element has order $1$? $\endgroup$ – bof Dec 4 '14 at 11:43
  • $\begingroup$ bof thank you. I have got my answer. $\endgroup$ – Anjan3 Dec 5 '14 at 5:56

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