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**NonCommAlg** define the function $\displaystyle f: G \longrightarrow G$ by $\displaystyle f(g)=g^{-1}T(g).$ now if $\displaystyle f(g_1)=f(g_2),$ then $\displaystyle T(g_1g_2^{-1})=g_1g_2^{-1}$ and thus $\displaystyle g_1=g_2,$ because we're given that $\displaystyle T(x)=x \Longrightarrow x = 1.$ so $\displaystyle f$ is injective and thus

surjective, because $\displaystyle G$ is finite. now let $\displaystyle x \in G.$ then $\displaystyle x=f(g)=g^{-1}T(g),$ for some $\displaystyle g \in G.$ thus $\displaystyle T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}.$ hence $\displaystyle G$ is abelian.