-1
$\begingroup$

This question already has an answer here:

Let $G$ be a group. By an automorphism of $G$ we mean an isomorphism $f: G\to G$ By an inner automorphism of $G$ we mean any function $\Phi_a$ of the following form: For every $x\in G$, $\Phi_a(x)=a x a^{-1}$. Prove that every inner automorphism of $G$ is an automorphism of $G$ which means I should prove $\Phi_a$ is isomorphism? any suggestion? thanks

$\endgroup$

marked as duplicate by Jack Schmidt, Amzoti, vadim123, Micah, Jim Jun 10 '13 at 3:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ hint: consider the set of all elements in your group. What happens when you multiply each element by "a"? Do you obtain any duplicates? $\endgroup$ – hasnohat Jun 5 '13 at 6:45
1
$\begingroup$

$\phi_a(xy)=a(xy)a^{-1}=axa^{-1}aya^{-1}=\phi_a(x)\phi_a(y)$

$\phi_a(x)=\phi_a(y)\implies axa^{-1}=aya^{-1}\implies x=y$

$\phi_a$ is also surjective since for each $y \in G $ there exists $x=a^{-1}ya$ s.t. $\phi_a(x)=y$

$\endgroup$
0
$\begingroup$

To prove that every inner automorphism is indeed an automorphism, you need to show that

  1. $\Phi_a$ is a homomorphism
  2. $\Phi_a$ is surjective
  3. $\Phi_a$ is injective (i.e $\ker\Phi_{a} = \{e\}$)

All three are straightforward if you know your definitions.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.