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I have a group $G$ of order $2p$, where $p>2$ and prime. The additional thing that I also know, that $\exists a\in Z(G)\mid O(a)=2$. I need to prove that G is abelian.

But first, before that, tell me please, isn't here (in my next direction of thinking any objection to the given information).

So, I look at two sub-groups of $G$ (the order isnt prime so they exist and their order devides the order of the group) of order $p$, $A,B\le G\mid O(a)=O(b)=p$

They are cyclic because of their order (is prime). And their elements are the total elements in $G$ (because they are sub-groups and there are totally $2p$ elements, and here I look at different sub-groups with $p$ elements each).

So left to prove that any $a\in A$ abelian with any $b\in B$. Now I think about the second data that given. The center of $G$ is in $G$, so there is $g\in G\mid g^2=e$. But must $g\in A$ or $g\in B$, but their order is $p$ and $2$ does not divide any of them. Isn't that an objection ? Otherwise, which of the step is wrong, what can't I conclude from what I wrote?

Michael Hardy
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Ilya.K.
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  • What does the information tell you about $G/Z(G)$? – Daniel Fischer Aug 23 '15 at 19:17
  • Oh, sorry, It look's like I've already asked that question, but I want to prove it without using $G/Z(G)$. I mean without looking at it. So the question is stil open here or in the old publication. How to prove that it's abelian without looking at the qoutient group as you mentioned. – Ilya.K. Aug 23 '15 at 19:22
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    The error in your post, or at least one error, is this: You have two subgroups of order $p$ and you conclude that the union has $2p$ elements. That would be true if the two subgroups were disjoint. Two subgroups cannot be disjoint... – David C. Ullrich Aug 23 '15 at 19:27
  • Yes, you right. I just though maybe to take one sub-group of order p and the other of order p+1, but then I dont know anything about the second, right? I cant tell nothing about it cylicity or union with first, or the intersection with the first, so I guess it does'nt lead me anywhere. So how do I show it? – Ilya.K. Aug 23 '15 at 19:36

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