Studying character theory of finite groups using Martin Isaacs' book "Character Theory of Finite Groups", I came across the following problem (exercise 6.4):
Let $G$ be a finite group and let $\mathscr{H}(G) = \{H \subset G \mid \exists \psi \in \operatorname{Irr}(H) \text{ such that } \psi{\uparrow}^G \in \operatorname{Irr}(G)\}$. Consider $\mathbb{H}(G) = \bigcap \mathscr{H}(G)$. Prove that, if $G$ is an $M$-group (that is, if every irreducible character of $G$ is induced by a linear character of a subgroup), then $\mathbb{H}(G)$ is abelian.
This is one of those problems where I don't really know how to proceed. I believe the easiest way to go about this is probably to show that every irreducible character of $\mathbb{H}(G)$ is linear. So let $\psi$ be one such character, and write $$\psi{\uparrow}^G = \sum_{i=1}^t a_i\chi_i$$
as a combination of irreducible constituents, where $a_i \in \mathbb{N}^*$. Computing both sides at $1$ yields $[G: \mathbb{H}(G)]\psi(1) \leq [G: \mathbb{H}(G)] \sum_{i=1}^t a_i$, where we used the fact that $G$ is an $M$-group to express each $\chi_i$ as a character induced from a linear character of a subgroup, and used the fact that said subgroup contains $\mathbb{H}(G)$. This means $\psi(1)$ is bounded by the number of irreducible constituents that appear in the decomposition of $\psi{\uparrow}^G$.
This is pretty much where I get stuck. I think Clifford's Theorem probably plays a role somewhere, but I couldn't extract any meaningful information using it… My main difficulty is that I know next to nothing about the characters of $\mathbb{H}(G)$ besides the property I proved above…
Is this a good approach? If so, how do I continue? And, if not, how to tackle this problem?
Thanks in advance!
PS: Does the characteristic subgroup $\mathbb{H}(G)$ defined above have a name? I couldn't find anything about it elsewhere…
Best Answer
Theorem Define $\mathscr{H}(G)=\{H \leq G: \varphi^G \in Irr(G) \text{ for some } \varphi \in Irr(H)\}$. Put $H(G)=\bigcap_{H \in \mathscr{H}(G)}H$. Then the following hold.
$(a)$ $Z(G) \subseteq H(G)$
$(b)$ If $G$ is an $M$-group, then $H(G)$ is abelian
$(c)$ If $G$ is solvable, then $H(G)$ is nilpotent of class $\leq 2$ (in other words $H(G)' \subseteq Z(H(G)))$.
Proof In the sequel, CTFG is the book Character Theory of Finite Groups of I.M. Isaacs. The theorem combines the solutions to Problems (6.4) and (6.5) from that book.
(a) Let $\varphi \in Irr(H)$ with $\chi=\varphi^G \in Irr(G)$. Then (see also similarly (5.11)Lemma, CTFG), $Z(\chi)=core_G(Z(\varphi)) \subseteq H$, hence (see also (2.28)Corollary, CTFG) $Z(G) \subseteq Z(\chi) \subseteq H$ for each $H \in \mathscr{H}$. This proves (a).
(b) Observe first that $H(G)$ is a normal subgroup of $G$ (this follows from (6.1)Lemma in CTFG), so $[H(G),H(G)]$, being characteristic in $H(G)$, is normal in $G$. Now assume $G$ to be monomial and let $\chi \in Irr(G)$, and $\chi=\lambda^G$ for some linear $\lambda \in Irr(H)$, $H \leq G$. Again (5.11)Lemma from CTFG asserts that $ker(\chi)=core_G(ker(\lambda))$, is the largest normal subgroup of $G$ contained in $H$. Now $H(G) \subseteq H$, so $H(G)' \subseteq H' \subseteq ker(\lambda)$, since $\lambda$ is linear. But $H(G)' \unlhd G$, so we must have $H(G)' \subseteq ker(\chi)$. However (see (2.21)Lemma in CTFG), $\cap_{\chi \in Irr(G)}ker(\chi)=1$, so $H(G)'=1$, that is, $H(G)$ is abelian.
(c) Let $N \unlhd G$ and $\varphi \in Irr(N)$. Clifford's Theorem yields $I_G(\varphi) \in \mathscr{H}(G)$. So $H(G) \subseteq I_G(\varphi)$, hence $\varphi$ is $H(G)$-invariant. Let $h \in H(G)$ and $n \in N$ and $\lambda \in Irr(N)$ a linear character. Then $\lambda(hnh^{-1}n^{-1})=\lambda^h(n)\lambda(n^{-1})=1$. So $[H(G),N] \subseteq ker(\lambda)$ for any linear $\lambda \in Irr(N)$, whence $\color{blue}{[H(G),N] \subseteq N' \text{ for any } N \unlhd G}$. Recall from (b) that $H(G)' \unlhd G$, hence plugging this into the blue formula we get $\gamma_3(H(G)) \subseteq H(G)'' \subseteq G''$. With induction one gets: $\gamma_{k+1}(H(G)) \subseteq G^{(k)}$. Hence, if $G$ is solvable, $H(G)$ must be nilpotent.
Finally, in general, if $X$ is a nilpotent group of class $k \geq 3$, then the lower central series looks like this: $\gamma_{k+1}(X)=1 \lt \gamma_{k}(X) \lt \gamma_{k-1}(X) \lt \cdots \lt X$ and $[\gamma_{k-1}(X),\gamma_{k-1}(X)] \subseteq \gamma_{2k-2}(X)$ (see for example D.J.S. Robinson, A Course in the Theory of Groups, 5.1.11(a)). But $\gamma_{2k-2}(X)=1$, since $2k-2 \geq k+1$. Hence $\gamma_{k-1}(X)$ is an abelian characteristic non-central (since $1 \lt \gamma_k(X)=[\gamma_{k-1}(X),X]$), subgroup of $X$.
So, if $H(G)$ would have a nilpotency class $\geq 3$, then the formula in blue would give $[H(G),\gamma_{k-1}(H(G)]=1$, whence $\gamma_{k-1}(H(G))$ centralizes $H(G)$, a contradiction. We conclude that $H(G)' \subseteq Z(H(G))$.$\square$