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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.

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agenor-andrade: cardinal of this group is prime number use this theorem (if |G|=P s.t P be prime number then G is cyclic group) –  Maisam Hedyelloo Feb 12 '13 at 18:58
    
@Maisam, the notation means the group is generated by those $5$ elements, not that those are its only elements. –  Gerry Myerson Feb 14 '13 at 0:46
    
It's not hard to find out what cyclic group it is by typing "Fibonacci group" into the web, but I'm not finding any sources that give a proof of the answer. –  Gerry Myerson Feb 14 '13 at 1:03
    
I gave a proof F(2,5) is abelian at: math.stackexchange.com/questions/22593/… –  user641 Feb 16 '13 at 5:05

3 Answers 3

up vote 5 down vote accepted

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 $Z_{11}$.

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Maybe you should add why $u$ is injective. –  1015 Feb 14 '13 at 13:13
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@julien : Please have a look now, where I use Derek Holts result that $F$ is commutative, and Tietze transformations. –  coffeemath Feb 15 '13 at 13:05
    
Nicely done, +1! –  1015 Feb 15 '13 at 13:10

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 $G=\langle a \rangle$.

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Thanks for the solution @DerekHolt. I utter to use the symbol $f$ instead of $ e$, because $e$ in some cases the drive denotes the group. –  Agenor Andrade Feb 15 '13 at 13:09

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 $\mathbb{Z}_n$

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I've tried doing it but could not. Can you give me some more advice? I have even tried using diagrams Van Kampem, but it was also useless. –  Agenor Andrade Feb 14 '13 at 20:21

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