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April 17th, 2012, 04:24 PM  #1 
Newbie Joined: Apr 2012 Posts: 16 Thanks: 0  Normal Subgroups Proof
Let N be a normal subgroup of the group G. If N is cyclic, prove that every subgroup of N is also normal in G.

April 17th, 2012, 05:02 PM  #2 
Newbie Joined: Oct 2011 Posts: 29 Thanks: 0  Re: Normal Subgroups Proof
If N is cyclic it's isomorphic either to or to . If it's isomorphic to , let f a homomorphism that is an isomorphism when restricted to N. Let M be a subgroup of N. . This is equal to . Lets assume it isn't and obtain a contradiction. If , then its product with itself would be always 1. then there would be a subgroup of order 2 of Z, which is impossible. If the product with itself ever difered from i.k, it would not constitute a subgroup. Try doing the finite case. 
April 17th, 2012, 08:01 PM  #3 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Normal Subgroups Proof
there's no need to distinguish between the cases, just pick a generator x of N. since H is a subgroup of N, if h is in H, h = x^k, for some integer k. for ANY g in G, we have ghg^1 = g(x^k)g^1 = (gxg^1)^k. since N is normal, gxg^1 is again in N, say gxg^1 = x^m, for some integer m. therefore: ghg^1 = (gxg^1)^k = (x^m)^k = x^(mk) = x^(km) = (x^k)^m = h^m. now h^m is in <h>, which is contained in H (H is closed under multiplication), therefore ghg^1 is in H, that is: gHg^1 is contained in H, that is: H is normal. 
April 17th, 2012, 09:46 PM  #4 
Newbie Joined: Apr 2012 Posts: 16 Thanks: 0  Re: Normal Subgroups Proof
Deveno, This looks great, this is what I came up with along with others. Thanks much! 

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