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Proof about inner automorphism of a group

ianchenmu

Member
Feb 3, 2013
74
Let $G$ be a group. Let $a ∈ G$. An inner automorphism of $G$ is a
function of the form $\gamma_a : G → G$ given by $\gamma_a(g) = aga^{-1}$.
Let $Inn(G)$ be the set of all inner automorphisms of G.
(a) Prove that $Inn(G)$ forms a group. (starting by identifying an appropriate binary operation.)
(b) Define $\varphi : G → Inn(G)$ by $\varphi(a) = \varphi_a$. Verify that $\varphi$ is surjective homomorphism and identify the kernel of $\varphi$.
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hint: For all $g\in G$:

$$(\gamma_a\circ \gamma_b)(g)=\gamma_a[\gamma_b(g)]= \gamma_a(bgb^{-1})=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\gamma_{ab}(g)$$

That is, $\gamma_a\circ \gamma_b=\gamma_{ab}$.
 

ianchenmu

Member
Feb 3, 2013
74
Hint: For all $g\in G$:

$$(\gamma_a\circ \gamma_b)(g)=\gamma_a[\gamma_b(g)]= \gamma_a(bgb^{-1})=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\gamma_{ab}(g)$$

That is, $\gamma_a\circ \gamma_b=\gamma_{ab}$.
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]?[/FONT]
 
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jakncoke

Active member
Jan 11, 2013
68
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ?
Define $\varphi : G → Inn(G)$ by $\varphi(a) = \varphi_a$.
isn't every element in Inn(G) of the form $f_a$, where $a \in G$? So for every element, $f_a \in$ Inn(G), there is a corresponding a which lives in G, so indeed the homomorphism $\varphi$ is surjective.

I think the more important question is whether $\varphi$ is even a homomorphism. which means for $a,b \in G$, $\varphi(ab) = f_{ab}$, by fernandos assertition this equals $f_a f_b = \varphi(a)\varphi(b)$. So indeed it a Homomorphism.

The kernel is all elements $a \in G$, such that $f_a = f_e$, the identity map, which would mean $f_a = a^{-1} x a = a^{-1} a x = x$, (note: $a^{-1}$ is also in the center if a is in the center) or basically elements which commutate with every element in G. or the center Z(G).
 
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Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ?
$\varphi$ is surjective by construction, that is if $\gamma_a\in\mbox{Inn } G$ then $\varphi(a)=\gamma_a$. On the other hand:

\begin{aligned}
\ker \varphi=&\{a\in G:\varphi(a)=id_G\}\\=&\{a\in G:aga^{-1}=g\;\forall g\in G\}\\=&\{a\in G:ag=ga\;\forall g\in G\}\\=&Z(G)
\end{aligned}

That is, $\ker \varphi$ is the center of $G$.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
And therefore, by the First Isomorphism Theorem:

$\text{Inn}(G) \cong G/Z(G)$.
 

ianchenmu

Member
Feb 3, 2013
74
$\varphi$ is surjective by construction, that is if $\gamma_a\in\mbox{Inn } G$ then $\varphi(a)=\gamma_a$. On the other hand:

\begin{aligned}
\ker \varphi=&\{a\in G:\varphi(a)=id_G\}\\=&\{a\in G:aga^{-1}=g\;\forall g\in G\}\\=&\{a\in G:ag=ga\;\forall g\in G\}\\=&Z(G)
\end{aligned}

That is, $\ker \varphi$ is the center of $G$.
why $id_G=g$? Is $g$ the identity of G? why not 1?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
why $id_G=g$? Is $g$ the identity of G? why not 1?
The identity element of $(\mbox{Inn }G,\circ)$ is the identity map $id_G(g)=g$ for all $g\in G$. So, $\varphi(a)=id_G$ is equivalent to say $\varphi(a)(g)=id_G(g)$ for all $g\in G$, or equivalent to say $aga^{-1}=g$ for all $g\in G$.
 

ianchenmu

Member
Feb 3, 2013
74
The identity element of $(\mbox{Inn }G,\circ)$ is the identity map $id_G(g)=g$ for all $g\in G$. So, $\varphi(a)=id_G$ is equivalent to say $\varphi(a)(g)=id_G(g)$ for all $g\in G$, or equivalent to say $aga^{-1}=g$ for all $g\in G$.
and is the binary operation of $Inn(G)$ the function composition?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661