An Exercise of Finite Groups Prove that the group $F(2,5)$ given by $$\left&lt;a,b,c,d,f \mid ab=c, bc=d, cd=f, df=a, fa=b\right&gt;$$ is a finite cyclic group.
Here is brief description of a proof that $G = \langle a \rangle$ and hence abelian. This was done by hand, not by computer, and is easy if you know how to do coset enumeration. I am going to rename your generate $f$ as $e$ (do you not like $e$ for some reason?).

Let $H$ be the subgroup $\langle a \rangle$ of $G$ and label the cosets $H$, $Hb$, and $Hd$ 1,2, and 3. Then we get the following chain of implications.

1a=1 by definition.
1b=2 by definition.
1d=3 by definition.
1c=2 from 1ab=1c.
2c=3 from 1bc=1d.
3e=1 from 1de=1a.
3b=1 from 3ea=3b.
3d=2 from 3bc=3d.
2e=2 from 2cd=2e.
3a=2 from 3de=3a.
1e=3 from 1ea=1b.
2d=3 from 1cd=1e.


So we have $1d=2d=3$, so $1=2$, and hence $b \in H = \langle a \rangle$, and it now follows easily that $G=\langle a \rangle$.
Hint: Observe that calling the elements $a_i,\;\;i=1,..,5$, then you have $a_ia_{i+1}=a_{i+2}$, and it goes around it when it reaches the end: try to find the unit of the group $1_F$, and then try to prove its isomorphic to some $\mathbb{Z}_n$