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Let $G$ be a finite group such that $p_1p_2\mid |G|$, where $p_1$ and $p_2$ are two primes. We know that there exists an irreducible character $\chi\in Irr(G)$ such that $p_1p_2\mid \chi(1)$. We know that $N$ is a normal subgroup of $G$ such that $(p_1p_2,|N|)=1$. Can we say that always there exists an irreducible character $\eta$ in $Irr(G/N)$ such that $p_1p_2\mid \eta(1)$? Thanks for your helps.

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    $\begingroup$ I assume you mean the two primes to be distinct (otherwise this clearly does not hold). My guess would be no. For example, we would otherwise have that whenever we had such a character, both those primes would divide the order of the derived subgroup, which does not seem like it should be the case. $\endgroup$ – Tobias Kildetoft Feb 20 '14 at 8:35
  • $\begingroup$ Thanks Tobias Yes I suppose that they are different. Is there any example? $\endgroup$ – BHZ Feb 20 '14 at 8:40
  • $\begingroup$ I was not able to find one after a quick look through groups of order $60$ and $120$ in GAP, but probably an example will need quite large order. $\endgroup$ – Tobias Kildetoft Feb 20 '14 at 8:42
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The answer is 'no'. A counterexample is given by $G=C_7\rtimes C_6$ with faithful action. If you induce any non-trivial character from $C_7$, the induction will be 6-dimensional irreducible. But of course there is no irreducible 6-dimensional character of $G/C_7\cong C_6$.

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