# Thread: cardinality proof

1. ## cardinality proof

Let G be a group and let $A \leq G$ be a subgroup. If $g \in G$, then $A^g \subseteq G$ is defined as

$A^g=\{a^g | a \in A \}$ where $a^g=g^{-1}ag \in G$.

Show that the sets $A$ and $A^g$ have the same cardinality.
Hint: sets $A$ and $A^g$ have the same cardinality if and only if there exists an injective and surjective map $f: A \to A^g$.

Attempt: So, in order to show that the mapping f is injective I have to show that

$f(a)=f(a^g)=f(g^{-1}ag) \iff a = a^g$ (for $a \in A$ and $a^g \in A^g$).
But how can I deduce that when I don't know what the function f is?

2. You don't. You have to define a specific f and show that that f is both injective and surjective. If you simply assert the existance of such an f, you are assuming what you want to prove.

Since $A^g= {g^{-1}ag |g\in G}$ how about trying $f(a)= g^{-1}ag$?

3. Originally Posted by HallsofIvy
You don't. You have to define a specific f and show that that f is both injective and surjective. If you simply assert the existance of such an f, you are assuming what you want to prove.

Since $A^g= {g^{-1}ag |g\in G}$ how about trying $f(a)= g^{-1}ag$?
To show that the mapping is injective:

$f(a)=f(a^g) \iff g^{-1}ag=g^{-1}a^gg$
$\iff agg^{-1} = a^g gg^{-1}$
$\iff a=a^g$

Is this correct?

To show that the mapping is surjective:

For any $a^g \in A^g$ let $a=g^{-1}a^gg$ (since $a^g = g^{-1}ag$).

$f(a)=g^{-1}(g^{-1}a^gg)g$

$= g^{-1}a^gg = a^g$

Is it right?