Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb Z/p^2\mathbb Z$.

share|improve this question
2  
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
2  
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
1  
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
add comment

2 Answers

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 $\Bbb{Z}/p^3\Bbb{Z} \times \Bbb{Z}/p^2\Bbb{Z}$.

share|improve this answer
add comment

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 $k$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.