Let $G$ be a finite abelian group of rank $n$ and $H\rightarrow G$ a central extension with cyclic finite kernel. Is it true that we can find a faithful representation $H\rightarrow {\rm GL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$?

I feel something like this must be true from the fact that $G$ should admit an irreducible projective representation into ${\rm PGL}_{k(n)}(\mathbb{C})$ where $k(n)$ only depends on $n$.


2 Answers 2


The answer is no.

According to Theorem 1.3 of this paper, if $H$ is a nilpotent group of class $2$ with cyclic commutator subgroup, then the minimal degree $m_\mathsf{f}(H)$ of a faithful complex representation of $H$ is given by $$m_\mathsf{f}(H) = \sqrt{|H:Z(H)|} + m_\mathsf{f}(Z(H)) - 1.$$ So, for example, the minimal degree of the group $$G = \langle x,y,z \mid x^n=y^n=[x,z]=[y,z]=1, [x,y]=z \rangle$$ of order $n^3$ is $n$, and $G/Z(G)$ is abelian of rank 2.

  • 6
    $\begingroup$ Actually, it is enough to evoke Jordan. If one takes the group of upper unipotent triangular $3\times 3$ matrices (Heisenberg) over $\mathbf{Z}/p\mathbf{Z}$, it has rank 2 but no abelian subgroup of index $<p$. Hence the dimension of its smallest faithful representation tends to infinity when $p\to\infty$. (There are explicit sharp bounds for the Jordan theorem, namely $(n+1)!$ for large $n$, which here are very far from the right bound. Actually Wedderburn directly yields that the smallest faithful representation has dimension $p$ and it's maybe easier than Jordan's older theorem.) $\endgroup$
    – YCor
    May 13, 2020 at 7:55

Here is an elementary proof of a related general fact. Let $H$ be any finite nilpotent group with $H^{\prime} = Z(H)$ cyclic of order $m$. Let $z$ be a generator of $Z(H)$. Then $\langle z \rangle $ has a (faithful) linear character $\lambda$ such that $\lambda(z)$ is a complex primitive $m$-th root of unity.

Note then that all irreducible constituents of ${\rm Ind}_{Z(H)}^{H}(\lambda)$ are faithful. For let $\chi$ be one such. Then by Frobenius recipirocity (and Clifford' Theorem), ${\rm Res}^{H}_{Z(H)}(\chi) = \chi(1)\lambda$, so that ${\rm Res}^{H}_{Z(H)}(\chi)$ is certainly faithful. On the other hand, if $\chi$ were not faithful, then ${\rm ker} \chi$ contains a minimal normal subgroup $M$ of $H$. Since $H$ is nilpotent, $M \leq Z(H)$ ( for $M \cap Z(H) \neq 1$ and $M$ is minimal), contrary to the fact that $M \leq {\rm ker} \chi$ and ${\rm Res}^{H}_{Z(H)}(\chi)$ is faithful.

Furthermore, ( as is well-known, and may be found in the character theory text of I.M. Isaacs for example), since $H$ is nilpotent, it follows that if $\theta$ is any faithful irreducible character of $H$, then $\theta$ vanishes identically outside $Z(H)$. For choose $a \in H \backslash Z(H)$ and choose $b \in H \backslash C_{H}(a)$ we have $[a,b] = w$ for some $1 \neq w \in Z(H)$. Then $b^{-1}ab = wa$. Hence $\theta(a) = \theta(b^{-1}ab) = \theta(wa)$. But by Schur's lemma, $w$ is represented by a scalar matrix in any representation affording character $\theta$, and the scalar, $\alpha$ say, is not $1$ as $\theta$ is faithful . Hence $\theta(a) = \theta(wa) = \alpha \theta(a) $, so that $\theta(a) = 0$.

Hence for our previous faithful irreducible character $\chi$ of $H$, the orthogonality relations yield $|H| = \sum_{ h \in H}|\chi(h)|^{2} = |Z(H)|\chi(1)^{2}$, so that $\chi(1) = \sqrt{[H:Z(H)]}.$


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