Suppose that $H = \{1, h, h^2, \ldots, h^{n-1}\}$ is a normal subgroup of a finite non-abelian group $G$ having order $n$. It is known that $H$ is cyclic with generator $h$. Let $c_G$ be the number of conjugacy classes of $G$. Let $$c_G=nc_{G/H},$$ where $G/H/ is the quotient group.
Let $G'$ be the derived (or commutator) subgroup of $G$. For each $1\leq i \leq n-1$, it is known that $$h^i\notin G'.$$ This condition implies that G has a set of $n-1$ (distinct) linear representations $\lambda_i$ for $1 \leq i \leq n- 1$ with $$\lambda_i(h^i)\neq 1.$$ Is the following statement true?
There exists a primitive $n$th root (let's say $\beta$) of unity such that $$\lambda_i(h^ig)=\beta \lambda_i(g) \ \text{for every}\ g\in G.$$ In other words, we have $$\lambda_1(h^ig)=\beta \lambda_1(g), \ \lambda_2(h^2g)=\beta \lambda_2(g), \ \text{and so on}.$$ I know that if $n$ is replaced with a prime number, then above statement is true. Please help.
You can prove the following statement:
Any character of degree $1$ (so $1$-dimensional representation) $\chi$ of a proper subgroup $H$ of a finite group $G$ can be extended to a character of $G$ if and only if $\chi(g)=1$ for any $g\in H\cap G’$.
I can give you a proof of this statement if you are interested in, let me know.
However, in your case $H\cap G’=\{1\}$ so the condition of the statement is always satisfied.
How many distinct $1$-dimensional representations has $H$? The number is exactly $|H^*|=|H|=n$. In particular, for any $h^i\neq 1$ there is at least one representation $\lambda_i$ of $H$ (and so also its extension on $G$) such that $\lambda_i(h^i)\neq 1$. So you have at least $n-1$ distinct representations of $G$ such that $\lambda_i(h^i)\neq 1$.