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I was trying to prove that $S_3\times Z_2$ is isomorphic to $D_6$.

Well there are a lot of methods to do so.

But I was queried that

If we know if two abelian groups have same number of elements of each order then they are isomorphic?

Also, do there exist a abelain and a nonabelian group satisfying the property but still not isomorphic?

So my doubt is -- is it true If both are nonabelian and have same number of elements of each order then they are isomorphism?

Shaun
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Sobhmatics
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  • If they are nonabelian, then they are abelian? I think that's not what you meant to write. But anyway I'm sure the question has been asked and answered here before. I think there are nonisomorphic nonabelian groups of order 16 with identical order profiles. – Gerry Myerson Nov 23 '20 at 06:40
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    See, for example, the answer I posted to https://math.stackexchange.com/questions/62331/three-finite-groups-with-the-same-numbers-of-elements-of-each-order in 2011. – Gerry Myerson Nov 23 '20 at 06:42
  • Are you still here? It would be nice if you were to engage with the answer and comments your question has generated. – Gerry Myerson Nov 24 '20 at 11:02
  • That was not my Question Gerry .The Results are well known For Abelian Groups .That is If Both Are Abelian and have same no of elements of each order then they Are isomorphic ..And If one of them is non Abelian the result is not true.. So my Question is if both are nonabellean having same no. Of elements of each order .. What can we say?? And I had seen your article earlier .But My Questions is here Is both are Non Abelian what can we say – Sobhmatics Nov 24 '20 at 12:11
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    Look at the last sentence in the body of the question you posted: "Is it true If both are NONAbelian And Have same no. Of elements of each order then they are Abellean." So you are asking whether it is true that if they are nonabelian then they are abelian. And I am urging you to edit that last sentence of your question into something that makes sense. And the answer to which I linked gives an example of two non-isomorphic nonabelian groups having the same number of elements of each order. – Gerry Myerson Nov 24 '20 at 21:33
  • See also https://groupprops.subwiki.org/wiki/Element_structure_of_groups_of_order_16#Order_statistics Nontrivial semidirect product of $Z_4$ and $Z_4$, and Direct product of $Q_8$ and $Z_2$; also, Small group $(16,3)$ and Central product of $D_8$ and $Z_4$. – Gerry Myerson Nov 24 '20 at 21:50
  • Please engage, Sobh. – Gerry Myerson Nov 25 '20 at 23:15
  • Sorry For That Gerry That was a typing mistake I am very very sorry for that. – Sobhmatics Nov 26 '20 at 09:26
  • Here my Question is We know if both are Abelian and have same no. Of elements of each order then they are ISOMORPHIC. And we know if one of them is nonabellean and other is abellean then they are nonisomorphic rather they have same no. Of elements of each order............But if both are nonabellean and have same no. Of elements of each order.. Can we say both are ISOMORPHIC???? – Sobhmatics Nov 26 '20 at 09:27
  • OK, thanks for editing. Now: do you understand that your question has been answered and answered and answered? That you have been shown several pairs of nonisomorphic nonabelian groups with the same number of elements of each order? – Gerry Myerson Nov 26 '20 at 10:54
  • Sorry For the inconvenience but I couldn't find such examples Gerry – Sobhmatics Nov 27 '20 at 04:32
  • There are examples in the link in my comment from 6:42 on 23 November. There are examples in the comment where I give the link to groupprops. There are examples at the links in the answer @Qiaochu posted. I cannot understand how you can look at all of this and still say you couldn't find any examples. They are there in great profusion. – Gerry Myerson Nov 27 '20 at 06:14
  • I try to help. You don't respond. I give up. – Gerry Myerson Nov 28 '20 at 11:16
  • No no . Actually I felt bad For My Mistake That's why I did not say anything .Sorry ..You did your best to me Thanks – Sobhmatics Nov 29 '20 at 12:56

1 Answers1

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No; there are many more groups than there are possible sets of orders. In this MO answer you can find a counting argument that makes this precise and in this MO answer you can find an explicit minimal counterexample of order $16$ (although I haven't checked to see if you can find a counterexample where both groups are nonabelian).

Qiaochu Yuan
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