Show that the group $G$ is Abelian if and only if every irreducible character of $G$ is linear.

$\textbf{The question is as follows:}$

The group $G$ is Abelian if and only if every irreducible character of $G$ is linear.

There is an identical question as this question here link to the question. But I cannot understand the solution for its sufficiency. So I decided to ask it again.

$\textbf{Some attempt:}$

$\rm (a)$ $(\Longrightarrow)$ For $G$ Abelian, every $g \in G$ and every representation $(\rho, V )$ give elements $\rho(g) \in Hom_G(V, V )$, since the $\rho(g)$ for different $g$ commute. If $V$ is irreducible, these $\rho(g)$ must all be given by scalar multiplication. Then any subspace of $V$ is an invariant subspace, implying the existence of sub-representations and thus a contradiction if $V$ is not 1-dimensional.

$(\Longleftarrow)$ We know that a character $\chi$ of $G$ is called linear if $\chi(1) = 1$. So our question turns to say that if a finite group has only 1-dimensional irreducible representations, then it is Abelian.

For to prove this we note that the number of irreducible representations is the number of conjugacy classes. Since the sum of the squares of the dimensions of the irreducible representations is the size of the group, if they're all one-dimensional, there are as many conjugacy classes as group elements, i.e. each group element is in a conjugacy class of its own. Hence the group is Abelian.

Can someone please let me know if my proof is correct or not correct?

And if it is not correct? Can you please edit it for me?

Many thanks!

• I'm fine with everything except the setence starting " If V is irreducible, ..." I'm just not following that. – Thomas Andrews Jan 19 '18 at 15:12
• @DietrichBurde Thanks! No I couldn't give what I wanted to say. Sorry about that. I will change it now. – Nikita Jan 19 '18 at 15:12
• Basically, you should be more explicit about referencing Schur's Lemma. But otherwise, the proof is fine. – Thomas Andrews Jan 19 '18 at 15:17
• @ThomasAndrews Many thanks! Can you please improve my proof? – Nikita Jan 19 '18 at 15:20