Group $G$ of order $p^2$: $\;G\cong \mathbb Z_{p^2}$ or $G\cong \mathbb Z_p \times \mathbb Z_p$

Question

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

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 09 '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 09 '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 09 '13 at 20:27
Does this answer your question? There exists only two groups of order $p^2$ up to isomorphism. — Martin Sleziak, Dec 21 '19 at 10:49

amWhy · Accepted Answer · 2013-12-30T13:27:54.603

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Hint:

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, $\gcd(p, p) = p \neq 1$.

edited Dec 30 '13 at 13:27

answered Jun 09 '13 at 20:05

amWhy

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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 09 '13 at 22:00
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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 09 '13 at 22:07
Nice additional notes! =1 – Amzoti Jun 10 '13 at 00:32
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I know i came up with this a little late, but there it is my proof: We hava that G is a p-group. If G has an element with order $p^2$ then G is cyclic and we're done. But if G doesn't have such element then G has an element with order $p$ (Cauchy theorem). Let $g\in G$ be such element, then has order p, and it's a proper subset of G, and for a Sylow theorem (can't remember wich one) we have that is a normal subgroup of G. Let $k\in G$ but not in , then the order of k divides the order of G, then |k|=p hence is a p-group then is a normal subgroup of G. ...cont... – Ana Galois Jun 10 '13 at 16:55
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...cont... Now i have that $\cap$={$e_G$} then their "product" is a subgroup of G, and its order is $p^2$, hence that product has to be G, then G is the direct product of & $\rightarrow$ G is isomorphic to $\mathbb Z_p\times\mathbb Z_p$. Now the only thing is that i'm not sure if i can use the direct product to proof this :S – Ana Galois Jun 10 '13 at 17:03
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That's fine, Ana. It's not all necessary, if you use the Fundamental theorem...but it does the job, and is well thought out! Nice work! – amWhy Jun 10 '13 at 17:05
@amWhy , i think that sylow theorems doesn't imply that the subgroup generated by $g$ is normal in this case , i think she means the corollary of sylow theorem , but also , it doesn't imply that, as $$ is not $p$-sylow subgroup of $G$ , so the only possible way which i can think of to prove that $$ is normal is using the fact that if $p$ is the least prime divides $|G|$ then any subgroup of $G$ with index $p$ is normal subgroup of $G$ , am i right ? – FNH Jun 17 '13 at 00:08
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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 00:11
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@MathsLover I meant to ping you when I posted the reply immediately above. – amWhy Jun 17 '13 at 00:17
@amWhy , "icing on the cake ", is this a expression ? what does it mean ? – FNH Jun 17 '13 at 00:29
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@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 00:31
@amWhy , i will tell you a secret " i think that publishing the secret here makes it not a secret ! ", when i was 10 , i was forced to learn English in school and we were asked to memorize our first three words in English " they were, cat , hat and rat :D " and i couldn't do that so i was punished from the teacher and this punishment was materialistic,so i felt that English language is this something which will make me pain so i didn't like it for along along along time till i was forced to like it as it's the best language to study Maths,but after that,i found that it's a beautiful language. – FNH Jun 17 '13 at 00:48
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@Maths Lover Teachers can be cruel. I'm sorry about that, for you! I wish I knew other languages. I do know Spanish well enough to read and write it, but I learned a lot about English, too, by learning Spanish...things I took for granted. – amWhy Jun 17 '13 at 00:51
@amWhy , i know English , Arabic and French . but not so good at french , in reality , i'm bad at french , but i like its way of pronouncing the words! , i really want to learn German , i think it's a very important language in Science nowdays – FNH Jun 17 '13 at 00:56

score 8 · Answer 2 · answered Jun 09 '13 at 20:04

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Hints:

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

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

answered Jun 09 '13 at 20:04

DonAntonio

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Geoff Robinson · Answer 3 · 2014-11-27T22:30:37.963

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.

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 09 '13 at 21:05

Group $G$ of order $p^2$: $\;G\cong \mathbb Z_{p^2}$ or $G\cong \mathbb Z_p \times \mathbb Z_p$

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