Why is $H(r, 4, r-1)\cong \Bbb Z_5$ for $r=2, 3\pmod{4}$? I'm due to start my PhD this October and I'll be working closely with $H(r, n, s)$, so a detailed answer aimed at that level would be great.
The Details.

Definition 1: Let $w=x_1\dots x_r(x_{r+1}\dots x_{r+s})^{-1}$. Define $H(r, n, s)$ by the presentation
  $$G_n(w)=\langle x_1, \dots, x_n\mid w, \theta(w), \dots , \theta^{n-1}(w)\rangle,$$ where $\theta$ is the map given by the permutation $(12\dots n)$ acting on $x_1, \dots, x_n$, for $r>s\ge 1$.

The following problem is from "On a class of finitely presented groups of Fibonacci type," by C. M. Campbell and E. F. Robertson. The proof of the result is said to be routine.
The Problem.

Why is $H(r, 4, r-1)\cong \Bbb Z_5$ for $r=2, 3\pmod{4}$?

Thoughts.
Obviously, we'll be using Tietze transformations to reduce the presentation to $\langle a\mid a^5\rangle$. (If there's any other way, let me know.) Other than that, I don't know where to start.

EDIT:
For the $H(r, 4, r-1)$ and $r=2$ case, we have that $$\langle a, b, c ,d\mid ab=c, bc=d, cd=a, da=b\rangle$$ and that $$ab=c\text{ and }cd=a$$ imply that $b^{-1}=d$ and, similarly, we have $a^{-1}=c$; we then have the following unique way to fill in the multiplication table:
$$\begin{array}{c|ccccc} 
 & e & a & b & c & d \\
\hline
e & e & a & b & c & d \\
a & a & \color{red}d & c & e & \color{red}b \\
b & b & \color{red}c & \color{red}a & d & e \\
c & c & e & \color{red}d & \color{red}b & a \\
d & d & b & e & \color{red}a & \color{red}c 
\end{array}.$$ Hence by inspection $H(2, 4, 1)\cong \Bbb Z_5$.

Note.
Here's a similar result:

$H(r, 6, r-1)\cong \Bbb Z_{13}$ if $r=3,4\pmod{6}$.

I suppose the proof of that would be similar to the answer to this question. I plan to tackle this one alone.
Please help :)
 A: Let's consider the case $r \equiv 2 \bmod 4$. So we have four generators $a,b,c,d$ and relations $(abcd)^kab = (cdab)^kc$ for some $k \ge 0$, and its three other cyclic permutations. This relation together with the third one, $(cdab)^kcd = (abcd)^ka$ imply $d=b^{-1}$, and similarly $c=a^{-1}$. So, eliminating $c$ and $d$, the first two relations become
$$(aba^{-1}b^{-1})^kab = (a^{-1}b^{-1}ab)^ka^{-1},\ \ \ (ba^{-1}b^{-1}a)^kba^{-1}=(b^{-1}aba^{-1})^kb^{-1}.$$
I had some difficulty making further progress, and I have the feeling there should be a quicker solution than the one below, but it seems to work.
Putting $u=(a^{-1}b^{-1}ab)^k$, these become $$abu=ua^{-1},\ \ \ bua^{-1}=aua^{-1}b^{-1}.$$
The first of these implies $uau^{-1} = b^{-1}a^{-1}$, and the second implies $bu = aua^{-1}b^{-1}a = aua^{-1}b^{-1}abb^{-1}$ which, since $u$ commutes with $a^{-1}b^{-1}ab$, equals $b^{-1}abub^{-1}$, so $ubu^{-1} = b^{-2}ab$.
But the equation $bua^{-1}=aua^{-1}b^{-1}$ can also be rewritten as 
$$ua^{-1}b^{-1}au^{-1} = a^{-1}b = (uau^{-1})^{-1}(ubu^{-1})^{-1}(uau^{-1}),$$
and then substituting in the expressions we have derived for $uau^{-1}$ and $ubu^{-1}$, gives $$a^{-1}b = (ab)(b^{-1}a^{-1}b^2)(b^{-1}a^{-1}) = ba^{-1},$$
so $a$ and $b$ commute, and hence $u=1$ and the equations reduce to $ab=a^{-1}$, $ba^{-1}=b^{-1}$, so $b=a^{-2}$ and $a^5=1$ and $|G|=5$.
I am going to leave the case $r \equiv 3 \bmod 4$ to you!
