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If a finite abelian group has order a power of a prime p,

then the order of every element in the group is a power of p.

Hi

I used Lagrange's theorem that order of element in Group (order of subgroup generated by element) must divide order of the group.

If G is order of power of prime, then order of element is power of prime.

But I can't understand whether this group is cyclic or not.

Also, Using Lagrange's theorem does not tell me whether every element has power of prime.

I think order of element does not necessarily have to be power of prime, as long as order of element divides power of prime.

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    $\begingroup$ If an integer divides $p^n$ for a prime $p$, then the integer is itself a power of $p$. $\endgroup$ – Alex G. Feb 23 '15 at 1:26
  • $\begingroup$ Just a suggestion: look at Abelian groups of order 8 (which are easy to play around with). $\Bbb Z_4 \times \Bbb Z_2$ and $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ might be of particular interest-compute the order of every element. $\endgroup$ – David Wheeler Feb 23 '15 at 7:20
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Every element has a prime-power order in your situation. This is because the only divisors of a prime power (the order of your group) are themselves powers of the prime.

The group may or may not be cyclic. For instance $\mathbb{Z}_4$ is cyclic, but $\mathbb{Z}_2\times \mathbb{Z}_2$ is not.

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  • $\begingroup$ How can we guarantee that every element has order of power of prime? $\endgroup$ – hho Feb 23 '15 at 1:29
  • $\begingroup$ If the group has order $p^a$ and every element order divides this group order, the only possible element orders are $p^0, p^1, p^2,\dots, p^a$ (since these are the only divisors of $p^a$). $\endgroup$ – paw88789 Feb 23 '15 at 2:05
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The order of $g\in G$ can also be viewed as the order of the subgroup $\langle g \rangle $. Which by lagrange divides $|G|=p^\alpha$

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  • $\begingroup$ Then it seems like this G is cyclic group since every element has order of prime power $\endgroup$ – hho Feb 23 '15 at 1:32
  • $\begingroup$ not necessarily. For example if the order of $G$ is $27$ the order of every $g\in G$ might be $3$. $3$ is a power of the prime $3$. For the group to be cyclic you would have to have an element of order $27$. $\endgroup$ – Jorge Fernández Hidalgo Feb 23 '15 at 1:34
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There seems to be some confusion here.

Let's say we have a group $G$, whose order is $p^n$, where $p$ is some prime number. Such a group is called a $p$-group, and there's a whole subfield of group theory devoted to $p$-groups.

In general, there's a limited amount of things you can say about a generic $p$-group. It could be cyclic, or abelian, or even non-abelian.

So, for your group, you can definitively say that every element must have order that's a power of $p$, as the only divisors of $p^n$ are integers of the form $p^k$, for some integer $k \leq n$. Only knowing that you have a $p$-group won't tell you much, until you learn more in-depth group theory (one elementary fact is that the center of a $p$-group has size at least $p$, but many more results are more in-depth).

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