Groups of a specific order from a difference set.

Question

So I am reading some surveys about Design Theory, and I am in over my head to a good extent. The text mentions a cyclic difference set in a group of order:

$$v=|G|=\dfrac{q^{d+1}-1}{q-1}.$$

Apparently, there exists a symmetric $(v,k,\lambda)$-design $PG_{d-1}(d,q)$ on $v$ points with $k=\frac{q^d-1}{q-1}$ and $\lambda=\frac{q^{d-1}-1}{q-1}$. This is all well and good, and I know $PG_{d-1}(d,q)$ is the design of points and (hyper)planes in the projective space $PG(d,q)$ over $GF(q)$, but what on earth would be the group here? $PG_{d-1}(d,q)$ is decidedly not a group.

To clarify, the difference set must come from a group of order $v$ given above. What group has that order? I thought of $PSL$, $PGL$, etc. but those all have slightly weirder group order formulas, and the group must be cyclic anyway.

What groups have order $v$?

Answer 1

These are Singer difference sets. Let $\mathbf{F}^*_{q^{d+1}}$ be the multiplicative group of $\mathbf{F}_{q^{d+1}}$, which is a cyclic group of order $q^{d+1}-1$ consisting of the nonzero elements of $\mathbf{F}_{q^{d+1}}$. If $\alpha$ generates $\mathbf{F}^*_{q^{d+1}}$, then $\alpha^{(q^{d+1}-1)/(q-1)}$ generates $\mathbf{F}^*_q$, which is therefore a subgroup of $\mathbf{F}^*_{q^{d+1}}$. The group you are looking for is $\mathbf{F}^*_{q^{d+1}}/\mathbf{F}^*_q$.