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Find the number of elements of order $p$, $p^2$, $p^3$, respectively in the group $\mathbb Z/p^3\mathbb Z \times \mathbb Z/p^2\mathbb Z$.

The answers are $p^2 − 1$, $p^4 − p^2$, $p^5 − p^4$, respectively. The key fact is that there are $p^3 − p^2$ elements of order $p^3$, $p^2 − p$ elements of order $p^2$, and $p − 1$ elements of order $p$ in $\mathbb Z/p^3\mathbb Z$.

How can I show the "why" of this answer? Similarly for $\mathbb Z/p^2\mathbb Z$.

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    Hint: answer the question for each factor separately. Since the factors are cyclic, you can start by looking at a generator.2012-04-27
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    Suggestion: Find the number of elements of exponent $1$; the number of elements of exponent $p$, the number of elements of exponent $p^2$, and the number of elements of exponent $p^3$. Then order $p$ is "exponent $p$ but not exponent $1$"; order $p^2$ is "exponent $p^2$ but not exponent $1$ or $p$", etc.2012-04-27
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    Did you come across the theorem that the number of elements of order $n$ of a cyclic group of order $n$ equals $\varphi(n)$?2012-04-27
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    Yes I think I got it! :)2012-04-27
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    @Arturo: Could you put that as answer so that this isn't hanging around as unanswered?2012-04-29

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