A question about cyclic

Question

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.

If $H$ is a normal subgroup of order $2$, what does that tell you about the centre of $G$? — Daniel Fischer, Jan 18 '14 at 23:21
What is $G/H$? Since $H \subset Z(G)$, $G/Z(G)$ must be a quotient of that. — Daniel Fischer, Jan 18 '14 at 23:38
You mean there are $p$ elements in $G/H$? Is that right? Actually I don't know why the order of $G/H$ is $p$, I am not sure about that even. — user121819, Jan 18 '14 at 23:42
For $G/H$ order check out http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29 — ir7, Jan 18 '14 at 23:50
@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? — Pete L. Clark, Jan 19 '14 at 03:16

score 5 · Answer 1 · edited Apr 13 '17 at 12:21

5

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jan 19 '14 at 00:22

Pete L. Clark

97,892

could you write a detailed answer, I just want to make sure I think it through – user121819 Jan 19 '14 at 01:52
@user121819: Well, what have you come up with so far? – Pete L. Clark Jan 19 '14 at 03:13

ir7 · Answer 2 · 2014-01-19T20:56:40.753

0

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/

edited Jan 19 '14 at 20:56

answered Jan 18 '14 at 23:40

ir7

6,249

Sorry, we have not learnt about Sylow's theorems so I really don't understand what's $K$ like. Could you explain it? – user121819 Jan 19 '14 at 00:15
1

Sylow's theorem seems unnecessary to bring up, really. – Pedro Jan 19 '14 at 00:17
I agree. We just need to show that $K$, subgroup of order $p$, exists. – ir7 Jan 19 '14 at 00:20
@user121819 this http://ysharifi.wordpress.com/2011/01/04/groups-of-order-2p/ seems to give an explicit (and better) proof. – ir7 Jan 19 '14 at 00:24
@Pedro Tamaroff I "fixed" it, I think. Best regards. – ir7 Jan 19 '14 at 01:25

A question about cyclic

2 Answers2