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March 1st, 2011, 12:04 AM   #1
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let G be p-group then every subgroup H, |H| = p, is normal s

let G be p-group then every subgroup H, |H| = p, is normal subgroup of G, Thank you,
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March 1st, 2011, 07:04 PM   #2
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Re: let G be p-group then every subgroup H, |H| = p, is norm

Hi again, johnmath.

How far have you gotten on the problem? It's generally a good idea to post whatever work you've done along with the problem statement so that you don't give the impression of just shifting onto others the burden of doing your homework problems. At the very least, let us know where you are getting stuck.

Do you remember anything about groups of order p (where I assume that p is prime)? Specifically, do they have a particularly simple structure? How do two groups of order p compare to one another? If they are both subgroups of the same group, how do they relate to one another within the group?

Keep in mind that we know something about the center of any p-group, and that Sylow's First Theorem tells us that p-groups have subgroups of all possible orders. Now, see if you can place one of your order p subgroups inside Z(G), and whether your answers to the questions in the previous paragraph lead you to a solution.

Best of luck!

-Ormkärr-
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March 1st, 2011, 09:47 PM   #3
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Re: let G be p-group then every subgroup H, |H| = p, is norm

I don't think, that the statement holds as stated. A counter example would be acting nontrivial on and taking G to be the semi direct product.

However what does hold are the following statemens:

1) Every nontrivial p-group G has at least one normal subgroup H of order p.
2) If G is a p-group and H is a subgroup of index p (i.e. (G:H)=p) then H is normal in G.

Please check your claim.

best
Peter
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March 3rd, 2011, 08:22 AM   #4
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Re: let G be p-group then every subgroup H, |H| = p, is norm

Peter,

Nice counter-example. I was trying to lead our OP on a journey of discovery by having him examine conjugates of the cyclic subgroup in a general setting, but I don't know if my hints were very clear.

I wanted to ask, however, is the same true for the other non-abelian group of order ?

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