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 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 two-step 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 $aKa^{-1}$ is subgroup of $H$ then you have to check $aKa^{-1} \neq \emptyset$ first what is obvious because $K$ is a subgroup of $G$ (thus $K \neq \emptyset$). Afterwards it satisfies to verify if $\forall x,y \in aKa^{-1}: x\cdot y^{-1} \in aKa^{-1}$ what Deveno showed you.

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