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February 15th, 2008, 04:21 PM   #1
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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!!!
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February 15th, 2008, 10:57 PM   #2
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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...
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February 16th, 2008, 04:33 AM   #3
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HeHe.....Hey it was a lot farther than I got, lol. Thanks!! This subject is a killer............
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February 16th, 2008, 10:01 AM   #4
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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.
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June 12th, 2008, 12:16 AM   #5
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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
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July 18th, 2008, 07:10 AM   #6
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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]
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August 13th, 2008, 03:01 PM   #7
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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.
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January 26th, 2015, 01:01 AM   #8
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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
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