Does an irr. rep. of finite $G$ have a basis of the form $\{hv : h \in H\}$ for $H$ subgroup of $G$?

Question

Let $G$ be a finite group and $V$ be a (complex vector space) representation of $G$. Consider the following three facts:

1. Whenever a $V$ is an irreducible representation of $G$, the dimension of $V$ must divide the order of $G$. See this stackexchange post: Proofs that the degree of an irrep divides the order of a group The proof uses the basis theory of algebraic integers.
2. Langrange's Theorem: Every subgroup of $G$ divides the order of $G$. (Though the converse is not true, in that if $d$ is an integer dividing the order of $G$, then there need not exist a subgroup of $G$ of size $d$.)
3. Whenever $V$ is a representation of a group $G$, and $v$ is any element of $V$, the set $\{ gv: g \in G \}$ spans a subrepresentation of $V$. Of course, it is natural to ask whether there is a subgroup $H$ of $G$ that does this more efficiently, and in particular if there is a subgroup $H$ of $G$ such that $\{ hv: v \in H\}$ also spans $G$. This appears to be the case for some simple examples of irreducible representations (e.g. the standard representation of $S_3$ is spanned by the action of the identity and a transposition on an eigenvector of a three cycle).

It is tempting to put these facts together and ask the following question:

Question: If $V$ is an irreducible representation of a finite group $G$, does there exist a subgroup $H$ of $G$ and an element $v$ of $V$ such that $\dim(V) = H$, and that $\{ hv: h \in H \}$ spans $V$?

If the answer to this question is "yes", this would give an alternative proof of Fact 1 above that circumvents the use of algebraic numbers.

Has anyone seen this question tackled in the literature? Is there a simple counterexample? If the answer is "no", can we amend the statement to say anything useful about spanning subsets of the form $\{ gv: g \in S\}$, where $S$ is any subset of $G$? How does this question pertain to the symmetric group $G = S_n$? Of course, perhaps the nontruth of the converse to Lagrange's theorem could suggest a "no" answer!

Anyway, I'd be grateful to hear an answer, reference, or any thoughts!

Answer 1

I think the answer to your question is no.

Say $V$ is an irreducible $\mathbb{C}[G]$-module, where $G$ is a finite group.

Let $H \leq G$, and suppose that $v \in V$ is such that $\{hv : h \in H\}$ is a basis for $V$. Then because the vector $\sum_{h \in H} hv$ is fixed by $H$, it follows that $H$ has a non-zero fixed point.

This suggests a way of finding counterexamples. You will have one if you can find a $V$ such that for all $H \leq G$, one of the following holds:

• $H$ has no non-zero fixed points on $V$.
• $|H| < \dim V$.

One such example is $G = Q_8$ (quaternion group) with $V$ the $2$-dimensional irreducible $\mathbb{C}[G]$-module.

Answer 2

The pair $(V, H)$ have this property iff the restriction $\text{Res}^G_H(V)$ of $V$ to a representation of $H$ is isomorphic to the regular representation of $H$. This is a very strong condition; it is equivalent to the condition that $\dim V = |H|$ and that the character $\chi_V(h)$ of $V$, when evaluated on $h \in H$, is $0$ for $h \neq e$.

So, we can find counterexamples by finding irreducible representations $V$ of finite groups $G$ (of dimension $\ge 2$) such that the character does not vanish on the non-identity elements of any subgroup of $G$ of order $\dim V$. As in testaccount's answer, the smallest counterexample appears to be the quaternion group $Q_8$: it has a unique subgroup of size $2$, namely its center $\{ \pm 1 \}$, and the character of the unique $2$-dimensional representation doesn't vanish on it. (Of course we can see pretty directly that this subgroup can't have the desired property, since it acts by a scalar on this representation.)

Interestingly it doesn't appear to be possible to produce counterexamples this way for the small alternating and symmetric groups, from an examination of their character tables.