If $F$ is the free group on $\{a,b\}$ and $N$ is the normal subgroup generated by $\{b^3, a^7, aba^{-2}b^{-1}\}$ I am trying to show that $F/N$ is a non abelian group on $21$ elements.

My concern is that I am not clear on how $N $ will look like. For example, it is intuitively clear that $a$ is not in $N$ but I am not sure how to prove this.

As per comments below I realize that it is somewhat difficult to understand what $N$ would look like. Is it possible to give a reference for the same?

  • $\begingroup$ I would just try to look at what order 21 groups exist, and try to construct a homomorphism to the one you think it is, then prove it is an isomorphism. $\endgroup$ – Carl Feb 6 '18 at 2:36
  • $\begingroup$ If you have the homomorphism from $F$ to the order 21 group it's easy to say things about $N$, since $N$ is precisely those words that are sent to the trivial element. $\endgroup$ – Carl Feb 6 '18 at 4:20
  • $\begingroup$ The most easy way to show that $a\notin N$ is clearly to find somehow, somewhere a group $G$ and two elements $A$ and $B$ verifying that $B^3=1_G$, $A^7=1_G$ and $AB=BA^2$ but $A\neq 1_G$. From the information you gave, the unique non-abelian group of order $21$ might be a good candidate for $G$. However, in general, there are (complicated) algorithms to take care of this, Magma magma.maths.usyd.edu.au/magma/handbook/text/812#9180 gives Computation with finitely presented groups by Sims as a reference. $\endgroup$ – Clément Guérin Feb 6 '18 at 4:41
  • $\begingroup$ @Clement thanks $\endgroup$ – Shahab Feb 6 '18 at 4:47
  • 1
    $\begingroup$ @Clément Well, in general, these algorithms do not work! In particular, it is in general undecidable if a presentation defines a finite group, or even defines the trivial group! (Of course, I give you a presenation and say that it is finite then everyone is happy because these algorithms work :-) ) $\endgroup$ – user1729 Feb 7 '18 at 14:16

Let $1$ be the identity element. We have the equivalences $$b^3=1,a^7=1,aba^{-2}b^{-1}=1$$ in $F$ that give us $F/N$.

Using $b^3=1$ and $b^2=aba^{-2}$ we can convert all powers of $b$ in a free product $a^{x_1}b^{x_1}...a^{x_n}b^{x_n}$ to $1$, so it looks like $a^{x_1}ba^{x_2}b...a^{x_n}b$.

Now we can manipulate the equivalence relations to get

$$bab=a^2$$ $$ba^2b=a^2ba^2$$ $$ba^3b=a^4ba^4$$

and so on (it's a little tedious), and use this to reduce the free product down to something of the form $a^{x_1}ba^{x_2}$, and then $a^xb^y$.

The idea is to show that you can reduce everything to $a^{\{0,1,2,3,4,5,6\}}b^{\{0,1,2,3\}}$ and conversely use this to generate $F/N$.

To answer your second question, note that multiplying by generators of $N$ always makes the length longer. This is because you won't ever have an element ending with $a^{\pm 1,\pm 7},b^{\pm 1, \pm 3}$ (unless you just multiplied by the inverse generator). Thus unless you start with $a$ you won't get $a$.

  • $\begingroup$ Yes, but why will a not be in N? $\endgroup$ – Shahab Feb 6 '18 at 2:48
  • $\begingroup$ I don't think that is necessary for the solution. But if $a\in N$ then $F/N$ would collapse down to only powers of $b$. $\endgroup$ – Akababa Feb 6 '18 at 2:54
  • $\begingroup$ I think you misunderstood the question. I want to understand what N (and not F/N) would look like and why it will not contain a. $\endgroup$ – Shahab Feb 6 '18 at 3:00
  • $\begingroup$ It's really the same thing as $F/N$ is just the cosets of $N$. $N$ is all the stuff you need to multiply $a^{x_1}b^{x_1}...a^{x_n}b^{x_n}$ by to get down to $a^{\{0,1,2,3,4,5,6\}}b^{\{0,1,2,3\}}$. $\endgroup$ – Akababa Feb 6 '18 at 3:07
  • $\begingroup$ I understand. Can you answer the question (forget about the quotient F/N) what is the structure of N and prove that a is not in N? $\endgroup$ – Shahab Feb 6 '18 at 3:28

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