An Exercise of Finite Groups Prove that the group $F(2,5)$ given by $$\left<a,b,c,d,f \mid ab=c, bc=d, cd=f, df=a, fa=b\right>$$ is a finite cyclic group.
 A: 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 $G=\langle a \rangle$.
A: This is to show that, given $F$ is commutative as shown by Derek Holt, one can show that $F$ is $Z_{11}$ by using Tietze transformations. Actually we only need the specific transformation of "removing a generator": If a generator has an expression in terms of the other generators, it can be removed after replacing it in the other relations.
As did Derek, I'll use $a,b,c,d,e$ for the generators. Starting with
$$ab=c,\ bc=d,\ cd=e,\ de=a,\ ea=b$$
we remove $e=cd$ by replacing it in the other four relations, to get
$$ab=c,\ bc=d,\ dcd=a,\ cda=b.$$
Next we remove $a=dcd$, replacing it in the other three relations, to get
$$dcdb=c,\ bc=d,\ cddcd=b.$$
Now for one more removal we replace $b=cddcd$ in the other two relations:
$$dcdcddcd=c,\ cddcdc=d.$$
So at this point we have two generators $c,d$ with the above two relations. 
Here is where heavy use of Derek's result that $F$ is abelian comes in, since we may rewrite the last two relations as
[1] $ccddddd=1,$
[2] $cccdd=1.$
Then from $cccdd=ccddddd$ comes $c=ddd$, which means we can do one more elimination of $c$ and end up with the single generator $d$, and then [1] becomes
$$(ddd)(ddd)ddddd=1,$$
i.e. $d^{11}=1$, while relation [2] becomes
$(ddd)(ddd)(ddd)dd=1,$
also $d^{11}=1.$
So using that $F$ is abelian and the Tietze transformation of replacing generators, we get that $F$ is indeed $Z_{11}$.
A: 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 $\mathbb{Z}_n$
