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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. – Brett Frankel Apr 27 '12 at 0:27
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. – Arturo Magidin Apr 27 '12 at 1:20
Did you come across the theorem that the number of elements of order $n$ of a cyclic group of order $n$ equals $\varphi(n)$? – Nicky Hekster Apr 27 '12 at 7:28
Yes I think I got it! :) – user30072 Apr 27 '12 at 22:42
@Arturo: Could you put that as answer so that this isn't hanging around as unanswered? – Tara B Apr 29 '12 at 11:05

A key fact of cyclic groups is that if $G$ is cyclic with order $n$, and $k|n$, then $G$ has only ONE subgroup of order $k$. So $\Bbb{Z}/p^3\Bbb{Z}$ has but a single subgroup of order $p$, namely $\langle[1]\rangle$. This subgroup contains ALL the elements of order $p$ (if there were another outside of it, it would generate another subgroup of order $p$).

Likewise, we have a sole subgroup of order $p^2$, which contains $\varphi(p^2) = (p-1)p$ generators, the other $p^2 - p(p-1) = p$ elements being the elements of order $p$ and the identity, $[0]$ (another way to see this is that the sole subgroup of $\Bbb{Z}/p^3\Bbb{Z}$ of order $p$ must likewise be the sole subgroup of order $p$ of the subgroup of order $p^2$, so subtracting out the $p-1$ elements of order $p$ and the identity, we have $p^2-p$ elements remaining, which must be of order $p^2$, the only possible order remaining). Again, there can be no other elements of order $p^2$ outside this subgroup, for if there were we would have at least TWO subgroups of order $p^2$.

Having accounted for all the elements of order $1,p$ and $p^2$, we see that we have $p^3 - p^2$ elements left, which must be of order $p^3$.

So, in finding (for example) the elements of order $p$ in $\Bbb{Z}/p^3\Bbb{Z} \times \Bbb{Z}/p^2\Bbb{Z}$, we note that either the element of the first factor is of order $p$ (and the element in the second group is of order $p$ or less), or the first element is the identity (of $\Bbb{Z}/p^3\Bbb{Z}$), and the second element is of order $p$. That gives us: $(p-1)p + p-1 = p^2 -1$ elements of order $p$ in $\Bbb{Z}/p^3\Bbb{Z} \times \Bbb{Z}/p^2\Bbb{Z}$.

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Hint. Count the elements of exponent $1$, of exponent $p$, of exponent $p^2$, and of exponent $p^3$. Note that the elements of order $p^i$ are the elements that are of exponent $p^i$ but not of exponent $p^{i-1}$.

(And note that an element $(g,h)$ of $G\times H$ is of exponent $k$ if and only if both $g$ and $h$ are of exponent $k$).

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