# Representation of central extension

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 $$n$$.

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.

• 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.)
– 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 $$\chi(1) = \sqrt{[H:Z(H)]}.$$