11
$\begingroup$

Given a finite group $G$, I am interested in finding a non-trivial proper subgroup $H$ of $G$ such that $\mathrm{Ind}_H^G\mathbf{1}$ contains all the irreducible representations of $G$, that is, permutation representation of $G$ on $G/H$ contains all the representations of $G$. I can see that if $G$ is abelian then this is not possible. Because, if $G$ is abelian and $H$ is such a subgroup then $\mathrm{Ind}_H^G\mathbf{1}$ will contain all $|G|$ many representations but $\mathrm{dim}(\mathrm{Ind}_H^G\mathbf{1})=[G: H]$ which is a contradiction.

Now, I am thinking about nilpotent groups. Given a finite nilpotent group $G$, is it possible to have such a subgroup? What about solvable groups?

I was also looking at this problem from another point of view. If $\mathrm{Ind}_H^G\mathbf{1}$ contains all the irreducible representations of $G$, then by Frobenius reciprocity we can say that for every irreducible representation $(\rho, V)$ of $G$, $V^{H}\neq {0}$. Hence, there is a one-one correspondence between the simple modules for the group algebra $\mathbb{C}[G]$ and the simple modules for the Hecke algebra $e_H\mathbb{C}[G]e_H$ (where $e_H=\frac{1}{|H|}\underset{h\in H}{\sum}h$). And I want to ask if there is some sort of classification of groups having subgroup with the property discussed above.

Also, given a compact connected Lie group $G$, when does it possess a closed subgroup $H$ such that $L^2(G/H)$ contains all the representations of $G$?

I posted this in MSE. Got no answer. That is why I am posting it here.

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ If such an $H$ exists, we may assume that $H$ has prime order. Also, $H$ is not normal. So for the quaternion group of order $8$, no such $H$ exists. I haven't checked whether the necessary condition, that $G$ has a non-normal subgroup of prime order, is sufficient. $\endgroup$ Commented Sep 27, 2024 at 8:18
  • 2
    $\begingroup$ If $G$ is nilpotent, then every irreducible character $\chi$ of $G$ has the form ${\rm Ind}_{X}(G)(\lambda),$ where $X$ is a subgroup of $G$ necessarily containing $Z(G)$, and $\lambda$ is a linear character of $X$. Mackey's formula then gives a criterion to check whether $H$ has fixed points on the induced representation: something like, there are fixed points if and only if $X \cap gHg^{-1} \leq {\rm ker} \lambda$ for some $g \in G.$ Not sure how helpful that is. $\endgroup$ Commented Sep 27, 2024 at 10:15
  • 3
    $\begingroup$ I do not have a good answer, but with Michael Magee, we have an example that we have found meaningful, for operator algebraic considerations: the pairs $(G,H)$ equal to $(\mathrm{SL}_4(\mathbf Z/N\mathbf Z),\mathrm{SL}_2(\mathbf Z/N\mathbf Z))$ for integers $N$ (or equivalently $\mathrm{SL}_4(\mathbf Z_p),\mathrm{SL}_2(\mathbf Z_p))$ for primes $p$) have the property that all nonzero representation over $\mathbf C$ of $G$ have a nonzero $H$-invariant vector. This is false for $4$ replaced by $3$, see arxiv.org/abs/2312.03220 $\endgroup$ Commented Sep 27, 2024 at 14:06
  • 1
    $\begingroup$ @LSpice in a corner, say the upper-left. $\endgroup$ Commented Sep 27, 2024 at 15:24
  • 1
    $\begingroup$ If $\sigma\in G$ has order $2$ and $H=\langle\sigma\rangle$ does not work, then there exists an irrep $\rho$ such that $\rho(\sigma)=-{\rm Id}$, so that $\sigma$ is central of order $2$ in $G/\ker\rho$. In particular, in every nonabelian simple group, such an $H$ always works. $\endgroup$ Commented Oct 1, 2024 at 21:50

1 Answer 1

Reset to default
11
$\begingroup$

Let us write ($*$) for the property that a subgroup $H$ of a finite group $G$ has nonzero fixed vectors in every representation of $G$.

Proposition 1: if $G$ is nilpotent, then no nontrivial subgroup $H$ of $G$ satisfies ($*$).

Proof: Assume such an $H$ exists. We may assume that $H$ is cyclic, generated by some nontrivial $h\in G$. Let $1=G_0 \triangleleft G_1\triangleleft \dots \triangleleft G_n = G$ be a central series. For some $i$, the element $h$ belongs to $G_{i+1}$ but not to $G_i$. Therefore, the image of $h$ in $G/G_i$ is nontrivial, but central. For every irreducible representation of $G/G_i$, the element $h$ has fixed points, but since it is central it acts trivially (Schur's lemma); therefore $h$ is trivial in $G/G_i$, a contradiction.

More generally, let $G$ be a finite group. We have the following simple obstructions:

  • If $N\triangleleft G$ is a normal subgroup and $H$ satisfies ($*$), then so does the image of $H$ in $G/N$.
  • A nontrivial central subgroup $H$ of $G$ never satisfies ($*$).
  • Combining these two, a subgroup $H$ whose image is nontrivial and central in some quotient $G/N$ never satisfies ($*$).

There is a converse for subgroups of order $2$.

Proposition 2: Let $H$ be generated by an element $h$ of order $2$. Assume that the image of $h$ in every quotient $G/N$ is either trivial or noncentral. Then $H$ satisfies ($*$).

Proof: Let $(V,\rho)$ be an irreducible representation of $G$ and let $N=\ker\rho$. If $h$ is trivial in $G/N$, then $h$ acts trivially on $V$ and therefore $H$ has nonzero fixed points in $V$. If $h$ is noncentral in $G/N$, then $\rho(h)$ has at least two distinct eigenvalues, so it has $1$ as an eigenvalue, i.e. $H$ has fixed points in $V$.

Corollary 3: Let $G$ be a nonabelian finite simple group, and let $H$ be a subgroup of order $2$. Then $H$ satisfies ($*$).

Proof: there is no nontrivial quotient, and $Z(G)$ is trivial. So Proposition 2 applies.

Corollary 4: Let $G = S_n$ for $n\ge 4$, and let $H$ be generated by a double transposition. Then $H$ satisfies ($*$).

Proof: $H$ is not central, and belongs to all the proper normal subgroups. So Proposition 2 applies.

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ This is a beautiful answer! I would just add a small comment regarding Corollary 3 for the non-group theorists among us: every nonabelian finite simple group has even order (this is the Feit-Thompson Theorem), hence has an element of order 2. $\endgroup$ Commented Oct 2, 2024 at 21:10
  • $\begingroup$ @Aurel In the proof of Proposition $2$, "If $h$ is noncentral in $G/N$, then $\rho(h)$ has at least two distinct eigenvalues". Why is that so? $\endgroup$ Commented Oct 3, 2024 at 6:37
  • 1
    $\begingroup$ @SoumyadipSarkar First, $h$ is noncentral in $G/N$ iff $\rho(h)$ is noncentral in $\rho(G)$. Now, if $\rho(h)$ had only one eigenvalue, then if would be a homothety (having finite order, it is diagonalisable), so it would commute with $\rho(h)$. $\endgroup$ Commented Oct 3, 2024 at 7:32
  • $\begingroup$ I meant to write that it would commute with $\rho(G)$. $\endgroup$ Commented Oct 3, 2024 at 8:21

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.