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I need to prove that $\mathbb{Z}_{p^r} \times \mathbb{Z}_{p^r}$ has $p^{r-1}(p+1)$ cyclic subgroups of order $p^r$.

My attempt is to find all elements $(a,b) \in \mathbb{Z}_{p^r} \times \mathbb{Z}_{p^r}$ of order $p^r$ and then divide them by $\phi(p^r)$. I know that in $\mathbb{Z}_{p^2}$ there are $p^2-p$ elements of order $p^2$. How can I count the number of elements in $\mathbb{Z}_{p^r}$ such that $p^r(a,b)=(0,0)$?

Can anyone help me?

Shaun
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Phi_24
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1 Answers1

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You are on the right track. An element $(a,b) \in \mathbb{Z}_{p^r} \times \mathbb{Z}_{p^r}$ is of order $p^r$ if and only if at least one of $a,b$ is of order $p^r$ in $\mathbb{Z}_{p^r}$. Since there are $\phi(p^r)$ such elements in $\mathbb{Z}_{p^r}$, by inclusion and exclusion, we get $$2\phi(p^r)p^r-\left(\phi(p^r)\right)^2$$ such $(a,b)$.

Now dividing by $\phi(p^r)$ as you suggest, will give the answer.

EDIT

When we count the elements, the expression $$2\phi(p^r)p^r$$ counts the elements where both $a$ and $b$ are of order $\phi(p^r)$ twice, so we have to subtract them.$$|A\cup B|=|A|+|B|-|A\cap B|$$

saulspatz
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  • Ok, I understand why $2\phi(p^r)p^r$, but can you please explicit why is it necessary to subtract $\phi(p^r)^2$? Thanks a lot for your help. – Phi_24 Jul 13 '21 at 15:17