Character of an irreducible representation vanishes outside of the commutator subgroup I need to prove the following proposition:

Suppose $\sigma$ is a degree -$d$ irreducible representation of a finite group $G$
, and it is the only one of that degree up to equivalence.
Then, for every $g\in G$ such that $g\notin G'$ we have that:
$\chi_{\sigma}(g)=0$

We were given a hint, which is to use the following lemma in order to aid for the proof of the proposition:

If $g\in G,g\notin G'$ then there is a degree-one representation
$\chi:G\rightarrow\mathbb{C}^{*}$ such that $\chi(g)\neq1$ .

Now, I have proved the lemma, but I cannot wrap my head around how to use it.
My guess, is construct another degree-$d$ representation using the tensor product, and use the fact that there is only one degree-$d$ representation up to equivalence, and
since they share the same character, deduce something from it.
I would really like help in here please.
Thanks
 A: Thanks to the kind commenters, I ended up proving the required, for the sake
of completeness I will post the solution for anyone else pondering on the same question:
Proof of the Lemma:
We know that $G'\triangleleft G$ so $G/G'$ is well defined.
Since $g\notin G'$ then $gG'\neq eG'$ .
Since every degree-one representation of $G$ is of the form $\pi=\phi\circ\psi:G\rightarrow\mathbb{C}^{*}$
where :
$\psi:G\rightarrow G/G'$
$g\mapsto gG'$
and $\phi:G/G'\rightarrow\mathbb{C}^{*}$
is an irreducible representation of abelian group therefore one-dimentional, then:
$\pi(g)=\phi(gG')\neq\phi(eG')=1$
Proof of the Proposition:
Let $\pi:G\rightarrow\mathbb{C}^{*}$ be a degree-one representation of $G$ such that $\pi(g)\neq1$.
$\tau:=\sigma\otimes\pi$ is also a degree -$d$ representation of $G$.
Let $g\in G$ where $g\notin G'$ .
$\chi_{\tau}(g)=\chi_{\sigma}(g)\chi_{\pi}(g)=\chi_{\sigma}(g)\pi(g)$
Since there is a unique irreducible representation of degree-$d$ up to equivalence then:
$\chi_{\sigma}(g)=\chi_{\tau}(g)=\chi_{\sigma}(g)\pi(g)$
$\Rightarrow\chi_{\sigma}(g)(1-\pi(g))=0$
for every $g\in G$ thus for $g\notin G'$ we have that $\pi(g)\neq1$ which means:
$1-\pi(g)\neq0$ thus $\chi_{\sigma}(g)=0$ .$\square$
