Show that $\psi_a:G\rightarrow G',a\in G,\psi_a(g)=aga^{-1}$ is homomorphism one to one and onto

Question

Let $G$ be a group and $\psi_a:G\rightarrow G,a\in G,\psi_a(g)=aga^{-1}$, I need to show that $\psi_a$ is homomorphism one to one and onto

It's not the same question like

"Is the conjugation map always an isomorphism?"

because here I need to show why $\psi_a$ is one to one and onto

Attempt:

$\psi_a(g_1g_2)=ag_1g_2a^{-1}$

$\psi_a(g_1)\psi_a(g_2)=ag_1a^{-1}ag_2a^{-1}=ag_1g_2a^{-1}$

$ \Rightarrow \psi_a $ is hom'

So far, so good. Now, one way to show that $\psi_a$ is one-to-one and onto is to find its inverse. Have you any idea how you could go about that? — Daniel Fischer, May 19 '15 at 12:33
Realize that $\psi(G)\subseteq G'$ is a subgroup of $G$. I agree with @DerekHolt. — drhab, May 19 '15 at 12:50
On your edit: showing that a homomorphism is bijective comes to the same as showing that it is an isomorphism (in this case even an automorphism). — drhab, May 19 '15 at 12:56

score 3 · Accepted Answer · answered May 19 '15 at 12:44

For one-to-one, suppose $\psi_a(g_1)=\psi_a(g_2).$ Hence $ag_1a^{-1}=ag_2a^{-1}$. Left multiplying each side by by $a^{-1}$ and right multiplying by $a$ yields $g_1=g_2$, hence $\psi_a$ is one-to-one.
For onto, suppose $g \in G$, then $a^{-1}ga \in G$ and $\psi_a(a^{-1}ga)=a(a^{-1}ga)a^{-1}=g$

score 2 · Answer 2 · answered May 19 '15 at 12:46

2

Hint:

If $\psi_a$ is homomorphism then so is $\psi_{a^{-1}}$. How are they related?

answered May 19 '15 at 12:46

drhab

151,093

Show that $\psi_a:G\rightarrow G',a\in G,\psi_a(g)=aga^{-1}$ is homomorphism one to one and onto

2 Answers2