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April 17th, 2012, 04:24 PM   #1
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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.
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April 17th, 2012, 05:02 PM   #2
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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.
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April 17th, 2012, 08:01 PM   #3
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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.
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April 17th, 2012, 09:46 PM   #4
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Re: Normal Subgroups Proof

Deveno,

This looks great, this is what I came up with along with others. Thanks much!
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