Question

Abstract algebra .....

Answer 1

http://www.math.niu.edu/~beachy/aaol/

This should help.

Answer 2

Let H \(\le\) G, and K \(\subset\) G such that:

\[K=\{x\in G~|~xax^{-1}\in H\}~\iff~a\in H\]
Show K \(\le\) G

Answer 3

coding messed up lol ...
\[K=\{x\in G~|~xax^{-1}\in H~\iff~a\in H\}\]

Answer 4

to be a subgroup we need to show 2 things, closure and inverses .... according to my book

Answer 5

closure:

Let n,m in G, and a in H. then if (nm) a (nm)' is in H; nm is in K

\[(nm) a (nm)'\]
\[(nm) a (m'n')\]
\[n~\underbrace{ (mam')}_{\in H}~n'\]
\[\underbrace{n~\underbrace{ (mam')}_{\in H}~n'}_{\in H}~~\therefore~nm\in K\]

Answer 6

inverses:

for some k in K, and a in H:\[kak^{-1}\in H\]
since H is a subgroup, it contains inverses such that:\[(kak^{-1})^{-1}\in H\]
\[k^{-1}~a^{-1}~[k^{-1})]^{-1}\in H~~\therefore~k^{-1}\in K\]

Answer 7

im thinking the inverse part in H may not be quite right ...

Answer 8

it wasnt on the final enyhows :)

Answer 9

looks right to me

Answer 10

The first part is incomplete. And the second is wrong (or maybe I'm mistaken??)

Answer 11

youre right, i just have difficulty with the definition for some reason

Answer 12

we have to show that the iff holds, so its bidirectional

Answer 13

yes. and that'll solve the problem with the first part. You just need to show it's bidirectional and it's pretty straight forward so no problems there. However, in the second part, you wrote \[(kak ^{-1})^{-1}=k a ^{-1} k ^{-1}\]

Answer 14

oh sorry I mean, what I wrote is the right equation.

Answer 15

yeah, which just shows that a' is in H

Answer 16

yeah. And that doesn't really get us anywhere. ;) So, what you need to do is to prove that \[x ^{-1}a x \in H \iff a \in H \] while assuming that \[x a x^{-1} \in H \iff a \in H \]

Answer 17

for some k in K, and a in H; kak' in H

does kak' = k'ak for all k in K ? assume it does and see if we get a contracition

kak' = k'ak

kkak' = ak

kkak'k' = a, only works if k is its own inverse :/

Answer 18

yes you are right about that. That is indeed a wrong assumption. So, assume a is in H. Put x'ax=t. This gives you a=xtx'. But a is in H. Which also means t will be in H. (using the given part as we have assumed that x is in K). Which means x'ax is in H. For the converse, assume x'ax is in H. i.e, x'ax=h. Which gives you a=xhx'. So as h is in H, xhx' is in H which means a is in H. So this proves both sides I guess. Which means x' is in K :) (pardon me if I missed something I'm in a hurry :P)

Answer 19

that looks about like the solutions ive come across :)

Answer 20

I see. This was the most straightforward way I guess :D