G is a group
for a in G define C_a = { x in G s.t. xa = ax}
prove that there is a one-to-one correspondence between the set of all the conjugates of a and the set of all the cosets of C_a

Question

G is a group
for a in G define C_a = { x in G s.t. xa = ax}
prove that there is a one-to-one correspondence between the set of all the conjugates of a and the set of all the cosets of C_a

zzr0ck3r · Answer

hint: use the fact that  x^-1*a*x = y^-1*a*y iff C_a*x = C_b*y

terenzreignz · Answer

What's a conjugate?

zzr0ck3r · Answer

for a in G its the for x^-1*a*x for x in G

zzr0ck3r · Answer

form*

kinggeorge · Answer

This makes intuitive sense, but I'm not immediately sure of the best way to approach it.

zzr0ck3r · Answer

do I need to show they are the say size?

zzr0ck3r · Answer

same size*

kinggeorge · Answer

That would be one way. To show that there are the same number of conjugates as there are cosets.

terenzreignz · Answer

Wait... conjugates is
$$\Large \left\{a\in G \quad \left|\quad xax^{-1} = a \quad \forall x \in G\right.\right\}$$

terenzreignz · Answer

^was a question

zzr0ck3r · Answer

if a is in G, a conjugate of a is any element of the form x*a*x^-1 where x is in G.

zzr0ck3r · Answer

so not all x

terenzreignz · Answer

ahh, okay understood, sorry :D

kinggeorge · Answer

Conjugates of $g\in G$ are $$\{a\in G|hgh^{-1}=a \text{ for some }h\in G\}$$

zzr0ck3r · Answer

much nicer:)

zzr0ck3r · Answer

number of cosets will be ord(g)/ord(C_a) I think?

kinggeorge · Answer

I think so.

terenzreignz · Answer

So we just have to find an injective map?

kinggeorge · Answer

Bijective map.

terenzreignz · Answer

It did just say one-to-one...

zzr0ck3r · Answer

1-1 correspondence means both

terenzreignz · Answer

ohh.  Okay.

zzr0ck3r · Answer

its silly wording

zzr0ck3r · Answer

well not silly but ...

anonymous · Answer

That hint really does the whole problem for you.

kinggeorge · Answer

agreed^

zzr0ck3r · Answer

I guess I just don't know what im trying to show. would it be the size?

terenzreignz · Answer

Maybe start with defining a set
$$\Large G_a = \left\{xax^{-1} \quad \left|\quad x\in G\right.\right\}$$

anonymous · Answer

If A is the set of all conjugates of a, then create the map:$$f: A\rightarrow G\big/C_a$$$$f(x^{-1}ax)=C_ax$$

anonymous · Answer

Or use teren's notation, thats much better. (instead of my A).

zzr0ck3r · Answer

ok so this is the map and I show that it is 1-1 and onto?

terenzreignz · Answer

And now, all that's left is to show that said mapping is both injective and surjective :D

terenzreignz · Answer

(I still don't know how to make that happen, but it's there)

anonymous · Answer

right.  and that hint does all of it except for the onto, which is easy enough.

anonymous · Answer

You also need to show that the map is well defined. (which again the hint shows).

terenzreignz · Answer

Showing it's onto may have been simpler than I thought...
Define an arbitrary coset of $\large C_a$  and show that there is a conjugate of a that will be mapped to said coset.

zzr0ck3r · Answer

so if I take and conjugate xax^-1 and another yay^-1 and assume f(one) = f(the other)
C_a*x = C_a*y iff our two elements of equal. thus its 1-1

to show onto take y in G/C_a  and this is where im a little lost

zzr0ck3r · Answer

I need to show that I can obtain y from some f(b) where b is a conjugate of a

anonymous · Answer

So given $$C_ay$$with $$y\in G$$can you think of a conjugate of a that maps to it?

anonymous · Answer

$$f(\_a\_)=C_ay$$what should go in the blank?

zzr0ck3r · Answer

sorry my wife walked in the door

zzr0ck3r · Answer

yay^-1

terenzreignz · Answer

yay :)

anonymous · Answer

right.

zzr0ck3r · Answer

ty sorry about that

anonymous · Answer

lol, no worries.

zzr0ck3r · Answer

and then as far as it being well defined I just need to show that ever element has an image I can do that:)

terenzreignz · Answer

More like you have to show that each element (in the domain) will have only one image.

anonymous · Answer

Good :) its important to show the function is well defined, you might want to put that before you show the 1-1 and onto.

zzr0ck3r · Answer

right

zzr0ck3r · Answer

@joemath314159 this might be silly but....
the way you have the function f(x^-1*a*x) 
but the book defines conjugates as x*a*x^-1
does this matter?
also I am a little confused on how to show that it is well defined.
what exactly do I need to show?

anonymous · Answer

in the big picture, there is no difference in$$x^{-1}ax$$versus $$xax^{-1}$$The only thing that will change are the details, but as long as you stick with only one, the main results will look exactly the same.  I would choose whichever way your book/class/professor defines conjugation.

zzr0ck3r · Answer

well the book says, when it defines conjugates x*a*x^-1 but when we use that hint the book says x^-1*a*x = y^-1*a*y iff C_ax = C_ay

zzr0ck3r · Answer

but we use the fact that our element is from the set of conjugates so I don't know how to show 1-1

anonymous · Answer

As far as a function being well defined, in general what you are trying to show is$$x=y\Longrightarrow f(x)=f(y)$$That statement looks trivial (and is in most cases), but in our setting, where we defined:$$f(x^{-1}ax)=C_ax$$we defined the output of the function based on on particular representation of the conjugate. If x is not equal to y and:$$x^{-1}ax=y^{-1}ay$$Then its clear that:$$f(x^{-1}ax)=f(y^{-1}ay)$$but:$$f(x^{-1}ax)=C_ax,f(y^{-1}ay)=C_ay$$Without the hint, we really dont know for sure that:$$C_ax=C_ay$$. If they didnt, our function would not be well defined.

anonymous · Answer

and to further highlight that the way you define a conjugate really doesn't change much, note that if we take $$x^{-1}ax$$to be a conjugate, let:$$g=x^{-1}$$Then you get:$$gag^{-1}$$The only difference is who you view as the inverse, the one on the left, or the one on the right.

zzr0ck3r · Answer

ok that's what I was thinking, but didn't know how to say it:)