Let $G$ be a group of order $2p$, where $p$ is an odd prime. If G contains a normal subgroup $H$ of order 2, show that $G$ is cyclic.

I was thinking to find a element and prove that it is the generator of $G$, but I cannot find that. could somebody give be some hints.

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    $\begingroup$ If $H$ is a normal subgroup of order $2$, what does that tell you about the centre of $G$? $\endgroup$ – Daniel Fischer Jan 18 '14 at 23:21
  • $\begingroup$ $H$ is in the centre of $G$ $\endgroup$ – user121819 Jan 18 '14 at 23:25
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    $\begingroup$ Right. So what are the possibilities for $G/Z(G)$? $\endgroup$ – Daniel Fischer Jan 18 '14 at 23:28
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    $\begingroup$ What is $G/H$? Since $H \subset Z(G)$, $G/Z(G)$ must be a quotient of that. $\endgroup$ – Daniel Fischer Jan 18 '14 at 23:38
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    $\begingroup$ @user121819: yes, if $\# G = 2p$ and $\# H = 2$, then $\# G/H = p$. You say you are unsure about this. Why? Do you know Lagrange's Theorem? $\endgroup$ – Pete L. Clark Jan 19 '14 at 3:16

Daniel Fischer has done a nice job giving helpful hints rather than a complete answer. I just wanted to mention that I think that the solution that he is nudging the OP towards uses this standard, elementary fact.

  • $\begingroup$ could you write a detailed answer, I just want to make sure I think it through $\endgroup$ – user121819 Jan 19 '14 at 1:52
  • $\begingroup$ @user121819: Well, what have you come up with so far? $\endgroup$ – Pete L. Clark Jan 19 '14 at 3:13

One direction would be to use the fact that there is (Cauchy's theorem) an element of order $p$ in $G$. The cyclic (order $p$) subgroup generated by it, call it $K$, is normal (has index 2). Then we can note that the two subgroups intersection is null and $G$ is generated by the their generators. If the two generators commute (they do, due to the normality of $H$), then $G$ is abelian, hence cyclic (order of the product of the two generators is $2p$).

See also the comments above (and the answer linked to them, posted above) and very explicit approach at http://ysharifi.wordpress.com/2011/01/04/groups-of-order-2p/

  • $\begingroup$ Sorry, we have not learnt about Sylow's theorems so I really don't understand what's $K$ like. Could you explain it? $\endgroup$ – user121819 Jan 19 '14 at 0:15
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    $\begingroup$ Sylow's theorem seems unnecessary to bring up, really. $\endgroup$ – Pedro Tamaroff Jan 19 '14 at 0:17
  • $\begingroup$ I agree. We just need to show that $K$, subgroup of order $p$, exists. $\endgroup$ – ir7 Jan 19 '14 at 0:20
  • $\begingroup$ @user121819 this ysharifi.wordpress.com/2011/01/04/groups-of-order-2p seems to give an explicit (and better) proof. $\endgroup$ – ir7 Jan 19 '14 at 0:24
  • $\begingroup$ @Pedro Tamaroff I "fixed" it, I think. Best regards. $\endgroup$ – ir7 Jan 19 '14 at 1:25

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