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If the order of $G$ is $p^2$ then how do I show that $G$ is isomorphic to $\mathbb Z_{p^2}$ or $\mathbb Z_p\times\mathbb Z_p$.

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It'd be nice to know what your ideas , self work are... – DonAntonio Jun 9 '13 at 20:04
Do you know the structure theorem for finite abelian groups? If so, show $G$ is abelian, and apply the theorem. It might be a little heavy handed. – Ben West Jun 9 '13 at 20:05
@Ben I do have that G is abelian, but i'm not quite sure if i can use that theorem, because we haven't even seen the definition of finite abelian groups in class yet, i mean not with the professor, i've read about it thought. – Ana Galois Jun 9 '13 at 20:14
@AnaGalois, you don't need the Fundamental Theorem for f.g. Abelian groups. If you can work out the hint in my answer you're then done, particularly because you already know $\,G\,$ is abelian... – DonAntonio Jun 9 '13 at 20:27
up vote 9 down vote accepted


Argue that $G$ must be abelian (why?)

Then use the Fundamental Theorem of Finitely Generated Abelian Groups to prove that any abelian groups of order $p^2$ must necessarily be isomorphic to one of the two groups $\mathbb Z_{p^2}$ or $\mathbb Z_p\times \mathbb Z_p$, which are non-isomorphic groups, since $\mathbb Z_m\times \mathbb Z_n \cong \mathbb Z_{mn} \iff \gcd(m,n) = 1$, and clearly, $\gcd(p, p) = p \neq 1$.

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You don't need for the group to be finite to use the fundamental theorem of finitely generated Abelian groups. Feel free to post follow up questions below answers. You can also use Lagrange to know that the only possible orders of subgroups other than the group of order $p^2$, which, since abelian, is necessarily cyclic, and the trivial group: the only subgroups if one exists, must have order $p$, and there must be more than one of those, since a subgroup of order p (prime) has only $p$ elements and is cyclic. What other order must another subgroup have, if a subgroup of order $p$ exists? – amWhy Jun 9 '13 at 22:00
That is, you have either that $G$ is cyclic (generated by one element), of order $p^2$, or you have that the order of the largest non-trivial proper subgroup of $G$ is $p$. This alone will not generate all of G. At least one more subgroup of order $p$ must exist. All you need to figure out is that these two subgroups must intersect trivially, and that we'll have 2 elements: a generator from each, that generate all of $G$. – amWhy Jun 9 '13 at 22:07
Yes, I think you're correct...Wish I could have been quicker with my answer to your most recent question! It's always icing on the cake to get "accepted." Nice question, though, and a natural question, too, that you asked. – amWhy Jun 17 '13 at 0:11
@MathsLover I meant to ping you when I posted the reply immediately above. – amWhy Jun 17 '13 at 0:17
@MathsLover It simply means I thoroughly enjoy answering questions (enjoying cake)...Even better is when cake has icing! (meaning, getting an answer accepted "feels really good", just like your getting 17+ upvotes on your " into function" question yesterday feels really good!) – amWhy Jun 17 '13 at 0:31


What are the possible orders of elements in $\,G\,$ ? And if there is no element of order $\,p^2\,$ in $\,G\,$ , then can you find two elements

$$1\neq x,y\in G\;\;s.t.\;\;\langle x\rangle\cap\langle y\rangle = 1\ldots?$$

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You can get by without the fundamental theorem of finitely Abelian groups in this case, once you know the group is Abelian. If there is only one subgroup of order $p,$ then consider the order of an element outside that subgroup. If there are two different subgroups $A$ and $B$ of order $p$. What is $A \cap B?$ What is $|AB|?$

It wasn't clear from what you wrote whether you already have proved for yourself that $G$ ( the whole group) a Abelian, or you know it for other reasons. Depending on what you have done so far on your course, this may or may not be straightforward. Have you proved that if $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is Abelian? Have you proved that a finite $p$-group has a non-trivial center? You didn't actually say that $p$ was a prime by the way, but I suppose that is clear from the context.

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yeah! we have prove that $G/Z(G)$ is cyclic then $G$ is abelian and that a finite p-group has a non-trivial center :D (I know that you are no mind-reader, p is a prime) – Ana Galois Jun 9 '13 at 21:05

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