Question

Hello,
Suppose a group G for * and that a belongs to G. Declare f: G to G and that f(x)=a^-1*x*a. a^-1 is the inverse element of a. f is an injective and surjective.
Now i have to prove that if b,c which belong to f are inversive then f(b) and f(c) are inversive.

Any help would be great. Thanks

Answer 1

It's a bit difficult for me to follow. Are you given that f is injective and surjective? Or do you want to prove that.

Answer 2

It's given, Already proven.

Answer 3

Have you learned the fact that any function that is bijective must have an inverse?

Answer 4

Yes, I have. But so far it didn't help me much.

Answer 5

Alright. Next part. What do you mean when you say that b,c belong to f, and are inversive?

Answer 6

Sorry, my bad. b and c belong to the group G and are inversive. Thank you, btw, for your help.

Answer 7

Do mean that if b and c are inverses of each other, then f(b) and f(c) are inverses of each other?

Answer 8

Yes

Answer 9

Great. Then look at \(f(b)f(c)\). This should equal e.\[
\begin{aligned}
f(b)f(c)&=(aba^{-1})(aca^{-1})\\
&=ab(a^{-1}a)ca^{-1}\\
&=a(bc)a^{-1}.
\end{aligned}\]Now, we are assuming that \(bc=e\), so we get\[aa^{-1}=e.\]Thus, \(f(b)f(c)=e\). Make sense?

Answer 10

You are putting * between each element, right?

Answer 11

Correct. I was merely being a bit lazy, but technically there is * between every element.

Answer 12

I need to leave now, but if you have any questions, just tag me in this post, and I'll be able to see it. Then I'll come back on later today, hopefully with an answer.

Answer 13

@KingGeorge, the transition between line 1 and 2 wasn't possible if G wasn't associative, right? It would be great if you could explain this transition. Thanks.

Answer 14

Yes, you do need associativity for that step. Fortunately, G is a group, so it has associativity.

Answer 15

Thank you again.