1. ## Automorphism - question

G - Finite Group
T - Automorphism of G
For all x in G, T(x) = x IFF x=1
Also, T(T(x))=x for all x in G.

Prove G is abelian.

Struggling a lot with this. I suspect $T(x) = x^{-1}$.
And once I establish this, proving G is abelian is simple.

But not going anywhere with the hunch of $T(x) = x^{-1}$.

2. Originally Posted by aman_cc
G - Finite Group
T - Automorphism of G
For all x in G, T(x) = x IFF x=1
Also, T(T(x))=x for all x in G.

Prove G is abelian.

Struggling a lot with this. I suspect $T(x) = x^{-1}$.
And once I establish this, proving G is abelian is simple.

But not going anywhere with the hunch of $T(x) = x^{-1}$.

define the function $f: G \longrightarrow G$ by $f(g)=g^{-1}T(g).$ now if $f(g_1)=f(g_2),$ then $T(g_1g_2^{-1})=g_1g_2^{-1}$ and thus $g_1=g_2,$ because we're given that $T(x)=x \Longrightarrow x = 1.$ so $f$ is injective and thus
surjective, because $G$ is finite. now let $x \in G.$ then $x=f(g)=g^{-1}T(g),$ for some $g \in G.$ thus $T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}.$ hence $G$ is abelian.
define the function $f: G \longrightarrow G$ by $f(g)=g^{-1}T(g).$ now if $f(g_1)=f(g_2),$ then $T(g_1g_2^{-1})=g_1g_2^{-1}$ and thus $g_1=g_2,$ because we're given that $T(x)=x \Longrightarrow x = 1.$ so $f$ is injective and thus
surjective, because $G$ is finite. now let $x \in G.$ then $x=f(g)=g^{-1}T(g),$ for some $g \in G.$ thus $T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}.$ hence $G$ is abelian.
I could never have guessed it - to define $f(g)=g^{-1}T(g)$ and then do it !!!