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 February 15th, 2008, 04:21 PM #1 Newbie   Joined: Nov 2007 Posts: 6 Thanks: 0 Conjugate Subgroups Help! For a subgroup H of G and a fixed element a ∈ G, let H^a = {x∈ G / axa^-1 ∈ H} 1) Show that H^a is a subgroup 2) Prove H^a={a^-1ya / y ∈ H} ..............I know that for the first one I need to show that closure holds, an identity exists, and inverses exist. But how to start is the hardest part I think. What two elements of H^a should I consider to show closure. Usually after I get by closure the inverses and identity follow, quite easily. Like I said the hardest part for me is the start. The 2nd one I am just clueless. Help!!! February 15th, 2008, 10:57 PM #2 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Alright, I'll try to give what help I can, although I'm pretty bad at algebra (read: do not take this as gospel... verify the hell out of it, because it's likely wrong!): closure: I had a great proof which relied on circular logic... instead: Assume x,y ∈ H^a Since axa^-1 (we'll call it x') and aya^-1 (y') are in H, we know x'y' is in H. e is in H^a: aea^-1 = aa^-1=e e is the identity of H^a: ey = (aea^-1)(aya^-1)=eaya^-1 = aya^-1; ye is the same. Existence of an inverse is trickier... maybe. If z,w ∈ G such that zw=wz=e, aza^-1awa^-1 =azwa^1=aea^-1=e Does that work? Or am I assuming more than I'm allowed to about z and w? **** I'm not so sure on the second part... I'll think about this. In fact, I'm not so sure on that first part... February 16th, 2008, 04:33 AM #3 Newbie   Joined: Nov 2007 Posts: 6 Thanks: 0 HeHe.....Hey it was a lot farther than I got, lol. Thanks!! This subject is a killer............ February 16th, 2008, 10:01 AM #4 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 I just wish I had a little more formal training in it. It's really interesting, but there's only so much you can pick up in your free time when you don't have any free time. On another note, thanks to that post I was doing algebra in my sleep... Group operations on music is always an interesting experience. June 12th, 2008, 12:16 AM #5 Newbie   Joined: Jun 2008 Posts: 3 Thanks: 0 Re: Conjugate Subgroups Help! i dunno if ne1 wud still read this page but i was just going through it and the question is quite easy ....1st i will show that 1 and 2 are equivalent then proving 2 to be a subgroup is elementary using our two step subgptest consider 1 ha={ x ? G|axa^-1? H} now, let x? ha then axa^-1? H (using def of ha) this implies axa^-1=h1 (h1? H) which implies x=a^-1h1a therefore, ha can be rewritten as ha={a^-1ha|h? H} which is precisely 1 therefore 1 and 2 are equivalent now you can easily prove that 2 is a subgroup of G using our two step test therefore 1 is also a subgp proved July 18th, 2008, 07:10 AM #6 Senior Member   Joined: Jul 2008 Posts: 144 Thanks: 0 Re: Conjugate Subgroups Help! 2 is intuitive,if you are good at set theory; so,H^a=a-1Ha f(x)=a-1xa:H->a-1Ha is an isomorphism from H to H^a so H^a is a group then is a subgroup of G. above is just my thought [qed] August 13th, 2008, 03:01 PM #7 Member   Joined: Aug 2008 Posts: 84 Thanks: 0 Re: Conjugate Subgroups Help! For 2, prove that they are subsets of each other (as the last poster hinted at). That aHa^-1 is a subset of H^a is obvious, since y is in H and H is a subgroup of G, then y is also in G. That H^a is a subset of aHa^-1 is as follows: let x be in H^a, then axa^-1 is in H so that axa^-1 = h, so then multiplying by a^-1 on the left and a on the right yields x = a^-1ha, which shows x is in aHa^-1 (this may not be immediately obvious, but aHa^-1 consists of all elements aha^-1, with h in H, so it holds for all elements of g, so we can have a^-1ha by replacing a with a^-1. January 26th, 2015, 01:01 AM #8 Newbie   Joined: Jan 2015 From: USA Posts: 1 Thanks: 0 Math has never been my strong suite, and I am currently having a difficult time solving the following problems. Would someone be kind enough to help me in solving these problems so I can check my own work. Thanks! A) 3log^3(6.47) B) xlog4=6-xlog25 Tags conjugate, subgroups Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pascal Uwoya Complex Analysis 1 August 15th, 2011 01:16 PM HairOnABiscuit Abstract Algebra 1 December 13th, 2010 05:59 AM srw899 Complex Analysis 1 February 23rd, 2009 04:14 PM mraevsky Linear Algebra 3 October 28th, 2007 10:53 PM payman_pm Abstract Algebra 6 April 8th, 2007 02:42 AM

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