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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 $Z$ or to $Z_n$. If it's isomorphic to $Z$, let f a homomorphism that is an isomorphism when restricted to N. Let M be a subgroup of N. $f(gmg^{-1})=f(g)kf(g)^{-1}$. This is equal to $kZ$. Lets assume it isn't and obtain a contradiction. If $f(g)kf(g)^{-1}=1$, 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 , 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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# let n be a normal subgroup of a group g. if n is cyclic, prove that every subgroup of n is also normal in g

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