# About the character degree of an extension

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 $15\mid \chi (1)$?

• Clifford's Theorem and Ito's theorem are the main tools to use here. – Geoff Robinson Aug 16 '16 at 13:38
• 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. – BHZ Aug 16 '16 at 13:46
• 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. – Geoff Robinson Aug 16 '16 at 13:52
• 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? – BHZ Aug 16 '16 at 15:38
• 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 $G$. – Geoff Robinson Aug 16 '16 at 16:05

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 $5$.

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 $G/N = A_5$.

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