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I try to find some relations for the irreducible character degrees of the extensions of the groups. For example:

Let $ G $ be a finite group of order $1800$ such that $ G $ has a normal subgroup of order $30$ and $ G/N $ is isomorphic to the alternating group $ A_5$. How we can prove that there is no irreducible character $\chi $ of $ G $ such that $15\mid \chi (1) $?

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  • $\begingroup$ Clifford's Theorem and Ito's theorem are the main tools to use here. $\endgroup$ – Geoff Robinson Aug 16 '16 at 13:38
  • $\begingroup$ I try but I could not prove this result since there exist two groups of order 30 such that their automorphisms have orders divisible by 5 or both 5 and 3. $\endgroup$ – BHZ Aug 16 '16 at 13:46
  • $\begingroup$ A group of order $30$ has a unique Sylow $3$-subgroup and a unique Sylow $5$-subgroup, and all its irreducible characters have degree $1$ or $2$ by Ito's theorem. $\endgroup$ – Geoff Robinson Aug 16 '16 at 13:52
  • $\begingroup$ Thanks for your omments. As you hint that $ N $ has a cyclic normal subgroup $ H $ of order 15 and so irreducible character degrees of $ N $ is 1 or 2. Therefore $ G $ has irreducible character degrees as $ b $ or $2b $ where $ b $ is a divisor of 60. But how we can get that $ b $ can not be equal to 15? $\endgroup$ – BHZ Aug 16 '16 at 15:38
  • $\begingroup$ There is some work to do. Note that $H \lhd G$, and that $H$ is centralized by each Sylow $3$-subgroup of $G$ and each Sylow $5$-subgroup of $G$. $\endgroup$ – Geoff Robinson Aug 16 '16 at 16:05
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The argument below is mainly group theoretic. I think it should be possible to give a more direct character-theoretic proof.

Let $H = O^{2}(N)$ which is cyclic of order $15$, and let $X = O^{2}(G).$ Then it is enough to prove that $X$ has no irreducible character of degree divisible by $15$, since $X \lhd G$ with $[G:X]$ a power of $2$, using Clifford's Theorem. Note that $H \leq Z(X)$ since $H \lhd X$ and ${\rm Aut}(H)$ is a $2$-group.

Let $P$ be a Sylow $5$-subgroup of $X$. Then $P$ is Abelian and $P \cap X^{\prime} \cap Z(X) \leq P^{\prime} = 1$ by elementary transfer. A similar argument works for the prime $3$. Hence $H \cap X^{\prime} =1$.

Now $XN/N \lhd G/N$ so either $X \leq N$ or $XN = G$. But if $X \leq N$ then $G$ is solvable, which is not the case. Hence $XN = G$ and $[G:X] = 2$.

Let $M$ be the terminal member of the derived series for $G$. Then $M \leq X$, so $M \leq X^{\prime}$. Hence $M \cap H = 1$ and $|M| \leq 60.$ Thus $M \cong A_{5}$ and $X \cong A_{5} \times H$, so all irreducible characters of $X$ have degree $1,3,4$ or $5$.

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Here is a more character-theoretic proof:

As mentionend in the comments, $N$ has a normal cyclic subgroup $H$ of order $15$, which is characteristic in $N$ and thus normal in $G$. By Clifford's theorem, we have $\DeclareMathOperator{\Irr}{Irr}$ $$ \chi_H = e \sum_{g\in [G:T]} \lambda^g \quad\text{for some } \lambda\in \Irr(H), $$ where $T=G_{\lambda}$ is the inertia group of $\lambda$. Since the Sylow $3$-subgroups and $5$-subgroups centralize $H$, the index $|G:T|$ divides $8$. Thus $15$ divides $e$, as $\chi(1) = e |G:T|$. By Frobenius reciprocity, $e = [\chi_H,\lambda] = [\chi, \lambda^G]$ and thus $$ 8 \cdot 15 = |G:H| = \lambda^G(1) \geq e\chi(1) \geq 15^2,$$ contradiction. (The last argument just reproves the well-known fact that $e^2\leq |T:H|$.)

Notice that I did not use the assumption $G/N = A_5$.

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  • $\begingroup$ Yhank you very much for nice answer. $\endgroup$ – BHZ Aug 17 '16 at 14:38

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