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April 17th, 2012, 09:45 PM   #1
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Subgroup proof

Let G be a group, a \in G, and let H, K be subgroups of G. Suppose K is normal with H and H is normal with G. Prove that aKa^{-1} is a subgroup of H, and that it is a normal subgroup. Note that aKa^{-1} = {aka^{-1}|k \in K}

*Note* must prove subgroup first via two-step check. Checking for 1.) closure, 2.) Inverses.
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April 18th, 2012, 01:21 PM   #2
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Re: Subgroup proof

let x,y be in aKa^-1.

thus x = aka^-1, for some k in K, and y = ak'a^-1 for some k' in K.

then xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1,

and since K is a subgroup kk'^-1 is in K whenever k,k' are, so xy^-1 is in aKa^-1.

thus aKa^-1 is certainly a subgroup of G.

what we need to do is show that aKa^-1 must be a subset of H.

note that K is a subgroup of H, hence aKa^-1 is a subset of aHa^-1. but H is normal in G, so aHa^-1 = H.

thus aKa^-1 is contained in H.
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April 18th, 2012, 10:30 PM   #3
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Re: Subgroup proof

Does xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1 prove both closure and inverses?
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April 19th, 2012, 12:54 PM   #4
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Re: Subgroup proof

Yes it does. If you need to check if a is subgroup of then you have to check first what is obvious because is a subgroup of (thus ). Afterwards it satisfies to verify if what Deveno showed you.
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