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Show that the set of all inner automorphisms of $G$, we will denote as $\mathrm{Inn}(G)$, is a subgroup of the set of all automorphisms of $G$, we will denote $\mathrm{Aut}(G)$, where we take $\mathrm{Aut}(G)$ to have the operation of function composition.

I have shown that $\mathrm{Inn}(G)$ is non-empty and that it is closed under composition of functions. I am now trying to show that the inverses of inner automorphisms are also inner automorphisms.

This is what I have so far in this regard:

$f:G \rightarrow G$ is an inner automorphism with $f(x)=c^{-1}xc$ for some fixed $c \in G$. If $a \in G$ then since $f$ is an isomorphism, it is also a bijection and so there is some $x,y \in G$ where $f(x)=a$ and $f(y)=c$. It follows $x=f^{-1}(a)$ and $y=f^{-1}(c).$ Hence $$f^{-1}(a)=f^{-1}(f(x))=f^{-1}(c^{-1}xc)=\left[ f^{-1}(c)\right]^{-1}f^{-1}(x)f^{-1}(c)=y^{-1}f^{-1}(x)y.$$

At this point I want $f^{-1}(x)=a$ but then $x=f(a)$ which seems suspicious as then $x=c^{-1}ac$ so then $f(x)=f(c^{-1}ac)=(c^{2})^{-1}ac^2=a$ so $(c^{2})^{-1}a=a(c^{2})^{-1}$ but we aren't assuming $G$ is abelian. Any tips?

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    $\begingroup$ Better notation will be helpful. Denote the inner automorphism $x \mapsto cxc^{-1}$ (I switched which one is the inverse for a reason) by $f_c$. $\endgroup$ – Qiaochu Yuan Mar 15 '13 at 5:43
  • $\begingroup$ @Wishingwell, that's a good comment by Qiaochu +1 . Check that if you don't do what he says then you get the product in $\,Inn\,$ reversed... $\endgroup$ – DonAntonio Mar 15 '13 at 6:28
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You can do that via this fact that $Inn(G)$ is non empty (You have shown this) and the fact that for all $g,h\in G$ $$f_gf_h=f_{gh}$$ so $$f_gf_g^{-1}=f_e$$ where $e$ is the identity of the group (You can see that $f_e$ is the identity of $Aut(G)$). so $$(f_g)^{-1}=f_{g^{-1}}$$ and from this we have $$(f_g)^{-1}f_h=f_{g^{-1}}f_h=f_{hg^{-1}}\in Inn(G)$$ Here we satisfy the condition in which any subset of a group should have to be a subgroup.

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  • $\begingroup$ Hello, Babak! I'm up now (morning time for me!) :-) $\endgroup$ – Namaste Mar 15 '13 at 13:01
  • $\begingroup$ @amWhy: .........;-) $\endgroup$ – mrs Mar 15 '13 at 14:29
  • $\begingroup$ Hello, dear! ;-) $\endgroup$ – Namaste Mar 15 '13 at 14:31
  • $\begingroup$ Thanks for your answer I appreciate it! $\endgroup$ – Frudrururu Mar 16 '13 at 11:13
  • $\begingroup$ @Wishingwell: Your very welcome. $\endgroup$ – mrs Mar 16 '13 at 18:04
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Inner automorphism, as a subset of automorphism is of course bijective. So you do not need to prove that. All you need is to prove it is closed under map composition, which is straight forward, contains identity, which is also easy, and exists an inverse, for which simply take $(f_c)^{-1}$ to be $f_{c^{-1}}$.

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You're overthinking it. How do you cancel out conjugation by $c$? Try conjugation by $c^{-1}$.

$$(c^{-1})^{-1}f(x)c^{-1}=cc^{-1}xc^{-1}c=x$$

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The easiest way to show that a function has an inverse is to write down the inverse. It's often the case that it's too hard to write down an inverse explicitly, but in this particular case it is easy.

What happens if you conjugate by $c$ and then conjugate by $c^{-1}$?

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I would aim at proving something more by considering the map $$ \varphi : G \to \operatorname{Aut}(G), \qquad g \mapsto (x \mapsto g x g^{-1}) $$ that maps $g \in G$ to the inner automorphism $f_{g} : x \mapsto g x g^{-1}$.

Show that $\varphi$ is a homomorphism of groups. It will follow that its image $\operatorname{Inn}(G)$ is a subgroup of $\operatorname{Aut}(G)$. Moreover, the first isomorphism theorem will tell you that there is an isomorphism $$ \frac{G}{\operatorname{ker}(\varphi)} \cong \operatorname{Inn}(G), $$ and one can verify that $\operatorname{ker}(\varphi) = Z(G)$ is the centre of $G$.

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