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

Prove the following properties.

$\rm (a)$ $~~$ The group $G$ is Abelian if and only if every irreducible character of $G$ is linear.

$\rm (b)$ $~~$ If $\chi$ is a character of an Abelian group $G$ then $[\chi; \chi] \ge \chi(1)$.

$\textbf{Some attempt for to prove first part:}$

$\rm (a)$ $~~$ $\Longrightarrow$

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 $\pi(g)$ 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.


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.

$\rm (b)$ $~~$ Since as what we saw in (a) for finite Abelian group, all its representation is of dimension 1 and there are $|G|$ such inequivalent representations. I was wondering if anyone could give a hint on how to show that $\chi(1) \le [\chi,\chi]$ at some cases.

Can someone please give me some hint on how to do it?



Any character $\chi$ is a nonnegative integral linear combination of irreducible characters. Thus you can write $\chi=\sum_i n_i\chi_i$ where each $n_i$ is a positive integer, and each $\chi_i$ is an irreducible character of $G$. By the first part, $\chi_i(1)=1$, so that $\chi(1)=\sum_i n_i$.

Computing $[\chi;\chi]$ by decomposing $\chi$ into irreducible characters as above, and using the fact that $[\chi_i;\chi_j]=\delta_{ij}$, gives $[\chi;\chi]=\sum_i n_i^2$, (when expanding out, by orthogonality we only have to worry about terms $[n_i\chi_i;n_i\chi_i]=n_i^2[\chi_i;\chi_i]=n_i^2$) and since $n_i^2\geq n_i$, the desired inequality is then clear. In fact, equality is achieved when $n_i^2=n_i$, that is, when $n_i=1$, so when $\chi$ is a sum of distinct irreducibles.

  • $\begingroup$ @Many thanks for your answer! Can you please write me the complete answer? Because I think in this case we just will get the equality and not $\chi(1) \le [\chi,\chi]$? Thanks for your time! $\endgroup$
    – Nikita
    Jan 16 '18 at 16:54
  • $\begingroup$ @Nikita Sure thing, I've expanded it a little bit. $\endgroup$
    – Ben West
    Jan 16 '18 at 21:21
  • 1
    $\begingroup$ Many thanks for your answer! $\endgroup$
    – Nikita
    Jan 16 '18 at 23:18

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