April 17th, 2012, 09:45 PM  #1 
Newbie Joined: Apr 2012 Posts: 16 Thanks: 0  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 twostep check. Checking for 1.) closure, 2.) Inverses. 
April 18th, 2012, 01:21 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  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. 
April 18th, 2012, 10:30 PM  #3 
Newbie Joined: Apr 2012 Posts: 16 Thanks: 0  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?

April 19th, 2012, 12:54 PM  #4 
Senior Member Joined: Jul 2011 Posts: 227 Thanks: 0  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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