# Homework Help: Is that mapping the homomorphism?

1. Apr 27, 2013

### NasuSama

1. The problem statement, all variables and given/known data

Let $G$ be a group and let $Aut(G)$ be the group of automorphisms of $G$.

(a) For any $g \in G$, define $\phi_{g}(x) = g^{-1}xg$. Check that $\phi_{g}(x)$ is an automorphism.

(b) Consider the map:

$\Phi:G \rightarrow Aut(G)$
$g \mapsto \phi_{g}$

Check that $\Phi$ is a homomorphism.

2. The attempt at a solution

Evaluate $\Phi$ at gh with arbitrary x, so:

$\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}g^{-1}xgh = \phi_{h}(\phi_{g}(x))$
$= \phi_{h} \circ \phi_{g} (x) = \Phi(h) \circ \Phi(g)$

But the operation is reversed for this situation. So this is considered to be an antihomomorphism.

2. Apr 27, 2013

### Fredrik

Staff Emeritus
In question (a), you probably meant "check that $\phi_g$ is an automorphism". Do you need help with (a)? You have only shown your work for (b).

At the far left, you need to write $\Phi(gh)(x)=\phi_{gh}(x)=$. The x appears out of nowhere in your calculation, and then it magically disappears at the end. So you need to do something similar at the end.

I agree that we end up with $\Phi(gh)=\Phi(h)\circ\Phi(g)$.

3. Apr 27, 2013

### Office_Shredder

Staff Emeritus
$$\left(\phi_{g} \phi_{h}\right)(x) = \phi_h \left( \phi_g(x) \right)$$
so you should check to make sure that multiplication is composition read from right to left

If this isn't what it's supposed to be then probably there's a type and the homomorphism is supposed to be
$$g \mapsto \phi_{g^{-1}}$$

4. Apr 27, 2013

### NasuSama

Nope, don't need help with part (a). Sorry to indicate that beforehand.