The group $G$ is Abelian iff every irreducible character is linear.

Question

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

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

Here there is a nice and short proof for this, but there are some assertions which I cannot understand.

There are identical questions here link to the question and link to the question. But the proof which comes is shorter.

Let $k$ be the number of conjugacy classes of $G$. Then $k = |G|$ if and only if $G$ is Abelian.

Now $|G| = \sum_{i=1}^{k} \chi_{i}^{2}(1)$ and $\chi_i(1)\ge 1$ for all $i$. It follows then $k = |G|$ if and only if $\chi_i(1) = 1$ for all $i$.

$\textbf{First:}$ Why it is true that " $k = |G|$ if and only if $G$ is Abelian?"

$\textbf{Second:}$ How did he conclude the result?

Can somone please explain it to me?

Many thanks!

Answer 1

All the conjugacy classes in an Abelian group have size $1$.

In a non-Abelian group, if $ab\ne ba$, then $ab$ and $ba$ are conjugate, so you have a conjugacy class of size $>1$. So the number of conjugacy classes is $<|G|$.

Answer 2

A group is abelian if and only if every element commutes with every other element.

If an element belongs to a conjugacy class of size $1$ it commutes with every other element because $b^{-1}ab=a$ implies that $ab=ba$

If an element belongs to a conjugacy class of size greater than $1$ you have an equation $b^{-1}ab=c$ with $a\neq c$ and $ab=bc\neq ba$ so the group is not abelian.

In an abelian group, therefore, all conjugacy classes have size $1$, and if the group is not abelian $ab=bc\neq ba$ implies that $a$ and $c$ are in the same conjugacy class, which doesn't have size $1$.

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For the second part, the arithmetic is done with characters, and characters belong to their associated representations. I assume you are working over $\mathbb C$ (an algebraically closed field).

Then, simply, it is a theorem of representation theory that the number of irreducible representations (characters) is equal to the number of conjugacy classes. So for an abelian group, this is the same as the number of elements.

Also the sum of the squares of the degrees of the representations (characters) is equal to the order of the group, so in the abelian case none of the representations can have order greater than $1$. And if there are fewer representations than elements of $G$, some representation must have order greater than $1$, and there will be a conjugacy class containing more than one element so the group will not be abelian.