Schur multiplier of $C_{2^n-1} \rtimes C_n$

Question

I am interested in computing the Schur multiplier of $C_{2^n-1} \rtimes C_n$ for any given $n \in \mathbb{N}$, where the action of $C_n$ is given as the automorphism of the Singer cycle $C_{2^n-1} \leq \operatorname{GL}_n(2)$ (so, in particular, this action is faithful).

I actually conjecture it is trivial for every $n$, and I have checked small examples I was able to build in GAP.

I was wondering if there is any general result that I am not aware of that could be applied in this case. If I am not mistaken, each group I am considering is a metacyclic group, which I have seen described as an "easy case" for calculating Schur multiplier. Nevertheless, so far I only found results when the two cyclic groups are coprime, an hypothesis which clearly does not hold in general here (for instance $n=6$ has $C_{63} \rtimes C_3$).

Answer 1

I won't delete my earlier answer, but here is a much simpler proof that your groups have trivial Schur Multiplier. This follows from the fact that they have balanced $2$-generator $2$-relator presentations $$\langle x,y \mid yxy^{-1}=x^2,y^n=1 \rangle.$$ Note that the two relations in this presentation imply $x = y^nxy^{-n} = x^{2^n}$, so $x^{2^n-1} = 1$, and hence the presentation defines your semidirect product $C_{2^n-1} \rtimes C_n$.

Answer 2

I think you are right in saying that these groups always have trivial Svchur Multiplier. Here is a very brief sketch of how I think this can be proved.

The metacyclic $p$-groups with trivial Schur Multiplier are listed in Section 5 of this paper by F. Rudolf Beyl.

I think, in the case when there is a prime $p$ dividing both $2^n-1$ and $n$, the Sylow $p$-subgroups of your semidirect product have the structure defined in groups of type I on that list, i.e. $G(p^a,p^b,p^{a-b}+1,0)$ with $p$ odd. (I have changed $m,n$ in the paper to $a,b$ to avoid clashing with your $n$.) So the Sylow $p$-subgroups have trivial Schur Multiplier and hence so does the whole group.

To see this, suppose that $n = p^bq$ where $p \not\vert q$.Then to get $p|2^n-1$, we must have $p|2^q-1$, and if $p^c$ is the highest power of $p$ dividing $2^q-1$, then the highest power dividing $2^n-1$ is $p^{c+b}$.

So a Sylow $p$-subgroup of $C_{2^n-1} \rtimes C_n$ is of the form $A \rtimes B$, where $A$ and $B$ are cyclic of orders $m := c+b$ and $n$, respectively. It can be seen that the centralizer in $A$ of $B$ has order $2^c$, which means that we do indeed have the group of type I, which is defined by the presentation $\langle x,y \mid x^{p^m}=y^{p^n}=1, yxy^{-1} = x^{p^c+1} \rangle$.