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A representation $\rho:G\to GL_k(\mathbb{F})$ is called fixed-point free if for every $1\neq g\in G$ and every $0\neq v\in \mathbb{F}^k$, $\rho(g)v\neq v$. Stated differently, it is a representation where $I-\rho(g)\in GL_k(\mathbb{F})\cup \{0\}$ for every $g\in G$. A group admiting such a representation is called a fixed-point free group.

I would like to find out which groups admit fixed-point free representations where also the sum of every 3 elements is invertible (or zero), i.e., $I_k+\rho(g_i)+\rho(g_j)\in GL_k(\mathbb{F})\cup\{0\}$ for every $g_i,g_j\in G$. Where can I find information about this problem?

Even more specifically, I'm interested in this property for representations of odd groups over fields with characteristic 2.

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you wrote "sum of every 3 elements", meaning "sum of every 2 elements and the identity"... – Dima Pasechnik Mar 30 '14 at 19:13
I think it is equivalent because one can always divide by one of the elements since they are invertible. The product is invertible/0 iff it was already invertible/0, and the representation property gives this new sum. But sum of every 2 elements and the identity is actually what I'm interested in. – A.B. Mar 30 '14 at 19:18
up vote 8 down vote accepted


If you intend to think about finite $G,$ then $|G|$ and the characteristic of $k$ must be relatively prime (for if the characteristic is $p,$ then an element of order $p$ would not act fixed point freely). Hence the representation is completely reducible. We might as well suppose that $k$ is algebraically closed. We might as well suppose that the representation is irreducible: any irreducible constituent is faithful by the fixed-point free condition.

Then all elements of order $3$ (if there are any) in $G$ are in $Z(G)$ for otherwise $I + x + x^{2}$ is not invertible for an element $x$ of order $3$ which is non-central. Then $G$ has a cyclic (possibly central) Sylow $3$-subgroup and a normal $3$-complement.

If the Sylow $3$-subgroup of $G$ is central, then it is a direct factor consisting of scalars. In that case, we might as well suppose that the group has order prime to $3$.

Given the remark at the end of the question, I assume now that the group $G$ has odd order.

Now $G$ is a Frobenius complement, and all Abelian subgroups of $G$ are cyclic. Hence $F(G)$ is cyclic and $G/F(G)$ is cyclic. Suppose that $G$ itself is not cyclic. Then the given representation of $G$ is induced from a representation $F(G)$ and is monomial.

I claim that $I + a+ b$ is either invertible or zero for $a,b \in F(G).$ For otherwise there
is $a \in F(G)$ of order $n,$ and there is a primitive $n$-th root of unit $\omega$ such that $1 + \omega + \omega^{j} = 0$ for some $j \neq 0,1.$ This forces $n = 3$ and $a$ to have order $3,$ (note that $b$ is a power of $a$ in this case),

The question reduces to considering monomial non-diagonal matrices $a,b$ of odd order such that $a+b$ has $-1$ as an eigenvalue. This splits into two case: $a^{-1}b$ diagonal and $a^{-1}b$ non-diagonal. I will return to this. An important case is when $a^{p}$ and $b^{p}$ are diagonal for some prime $p.$

Consider the eigenvalues of $I + c$ for $c$ a non-diagonal monomial matrix with $c^{p}$ diagonal of odd order , and $c$ of $p$-cycle shape. Then $(c+I)^{p} = \lambda I$ for some odd order root of unity $\lambda.$ The eigenvalues of $c$ are therefore of the form $\lambda \omega -1$ for some $p$-th root of unity $\omega.$ This can only be of the form $-\mu$ for some odd order root of unity $\mu$ if $p = 3.$

If $a^{-1}b$ is diagonal, again the only way that that $I + a^{-1}b$ can have $-\mu$ as an eigenvalue for some odd order root of unity $\mu$ is when that order is $3.$

We may conclude that there are no restrictions (apart from being a Frobenius complement) on the normal $3$-complement of $G.$ I omit the details but in general, when the group $G$ is a Frobenius complement of odd order, the extra condition of $I + a + b$ being invertible or zero imposes no further restriction. For $G$ necessarily has cyclic Sylow $3$-subgroups and a normal $3$-complement, and all elements of order $3$ are necessarily central in an odd Frobenius complement (eg by Hall-Higman Shult type results).

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Why would 3 even divide $|G|$ ? And why must the normal 3-complement be Abelian? – Dima Pasechnik Mar 30 '14 at 20:20
I have to repeat @DimaPasechnik 's question - why does there exist an irreducible representation of dimension 3? I might take the quaternions for example (they do not interest me since this is an even group, but SmallGroup(275,1) on GAP also seems problematic). Another thing: notice that I allow for the sum to be 0, just not a non-zero singular matrix. – A.B. Mar 31 '14 at 5:13
@Dima Pasechnik: Sorry, I misread the question- for some reason, I thought that the question was about $3$-dimensional representations. – Geoff Robinson Mar 31 '14 at 8:55
So basically, if I work with odd fixed-point free groups then I'm safe? – A.B. Mar 31 '14 at 10:57
I believe that is the case – Geoff Robinson Mar 31 '14 at 10:59

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