4
$\begingroup$

Fix some $n \geq 1$ and some prime $p$. I'm looking for finite $p$-groups $G$ and finite-dimensional complex representations $V$ of $G$ with the following two properties:

  1. The abelianization of $G$ has rank $n$; for instance, it could be $(\mathbb{Z}/p)^n$.

  2. For all nonidentity $g \in G$ and all nonzero $v \in V$, we have $g(v) \neq v$. In other words, $1$ is not an eigenvalue for the action of $g$ on $V$.

Here are the cases I know how to deal with:

  1. For $n=1$ and an arbitrary prime $p$, you can take the cyclic group of order $p$ and the $1$-dimensional representation where the generator acts as rotation by $2\pi/p$.

  2. For $n=2$ and $p=2$, you can let $G$ be the finite quaternion group and $V$ be the unique $2$-dimensional irreducible representation of $G$.

Improve this question
$\endgroup$
7
  • $\begingroup$ A $p$-group satisfying condition 2 is is indeed cyclic when $p$ is odd, and cyclic or generalized quaternion when $p = 2$. This can be found in many texts. $\endgroup$
    – Geoff Robinson
    Commented Feb 26, 2016 at 21:35
  • $\begingroup$ @GeoffRobinson: Can you suggest such a text? Thanks! $\endgroup$
    – Svetlana
    Commented Feb 26, 2016 at 21:47
  • $\begingroup$ Aschbacher's text "Finite Groups" should contain the necessary background. $\endgroup$
    – Geoff Robinson
    Commented Feb 26, 2016 at 21:57
  • 2
    $\begingroup$ The point is that if an elementary Abelian $p$-group $A$ of order $p^{2}$ acts faithfully as a group of linear transformations on a finite dimensional vector space $V$ over a field of characteristic different from $p$, then we may write $V = U \oplus W$, where $U$ and $W$ are both $A$-invariant and $A$ acts trivially on $U$ and without any non-zero fixed vectors on $W$. We may extend the ground field and assume that $A$ acts by diagonal matrices on $W$. Then for any $u,v \in A$, we see that $uv^{-i}$ must have a fixed vector on $W$ for some $0 \leq i \leq p-1$. Then 2 forces $u = v^{i}$. $\endgroup$
    – Geoff Robinson
    Commented Feb 26, 2016 at 23:22
  • 2
    $\begingroup$ Hence $A$ must be cyclic. The only finite $p$-groups which contain no elementary Abeliann $p$-subgroup of ordr $p^{2}$ or more are cyclic if $p$ is odd, and cyclic or generalized quaternion when $p = 2$. $\endgroup$
    – Geoff Robinson
    Commented Feb 26, 2016 at 23:29

1 Answer 1

Reset to default
3
$\begingroup$

(This question has been answered in comments by Geoff Robinson:)

You have already found (almost) all examples there are, namely cyclic $p$-groups and generalized quaternion groups in case $p=2$. The point is that every abelian subgroup of such a group $G$ is cyclic: Suppose $A\leq G$ is abelian. Let $\lambda \in \operatorname{Irr}(A)$ be a linear constituent of the character of $V$. Then every element in $\ker(\lambda)$ has non-zero fixed vectors in $V$. Thus $\ker(\lambda)=1$ and $A$ is cyclic.

By (say) Theorem 6.11 in Isaacs' text Finite Group Theory, a $p$-group with all abelian subgroups cyclic is cyclic or generalized quaternion. (This result should be in many other books on finite group theory.)

Finite groups ($p$-groups or not) which admit a complex representation such that Condition 2. holds are exactly the Frobenius complements (see here).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .