# Proving $G := \langle a, b, c \mid abc^{-1}a^{-1}, bcb \rangle$ is not isomorphic to $H := \langle a, b \rangle$

I'm trying to prove that $$G := \langle a, b, c \mid abc^{-1}a^{-1}, bcb \rangle$$ is not isomorphic to $$H := \langle a, b \rangle$$.

If they are isomorphic, then their abelianizations $$G/[G, G] = \langle a, b, c \mid abc^{-1}a^{-1}, bcb, abc^{-1}a^{-1}, bcb, aba^{-1}b^{-1}, bcb^{-1}c^{-1} \rangle$$ and $$H/[H, H] = \langle a, b \mid aba^{-1}b^{-1} \rangle$$ are isomorphic. Writing those groups as $$\langle a, b, c \rangle / N$$ and $$\langle a, b \rangle /M$$ for ease of notation, we can write any element in $$G/[G, G]$$ as $$a^nb^mc^kN$$ due to the words $$aba^{-1}b^{-1}, bcb^{-1}c^{-1}$$ and then as $$a^nc^{m + k}N$$ due to the word $$abc^{-1}a^{-1}$$. I don't see how we proceed from here.

• The chances are that the isomorphism problem for finitely presented groups is undecidable. – Shaun Apr 7 at 12:59
• @Shaun Well, that sucks. Does that go only for the problem in full generality, or is it (probably) impossible for any 2 finitely presented groups to prove if they are isomorphic or not? – Pel de Pinda Apr 7 at 13:02
• If $abc^{-1}a^{-1}=e$ then $bc^{-1}=e$, so $b=c$. But then we have that $c^3=e$, hence the groups are not isomorphic. – lulu Apr 7 at 13:08
• I mean that there is (probably) no general technique. No Turing machine, most likely, is capable of halting, after being given any two finite presentations, whenever the two define isomorphic groups. That's not to say that your question is impossible to solve; rather, that it is very difficult and has the potential of being impossible. – Shaun Apr 7 at 13:08
• Since the generator $c$ of $G$ can be eliminated, $G$ is generated by $a$ and $b$, so $H = \langle a,b \rangle = G$. So the groups are isomorphic. – Derek Holt Apr 15 at 12:08

You are almost done:

$$H/[H,H]\cong\mathbb{Z}^2$$

In $$G/[G,G]$$, we have $$1=abc^{-1}a^{-1}=bc^{-1}$$ so $$b=c$$. Therefore $$1=bcb=b^3$$ so $$G/[G,G]\cong\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$$

• This (i.e., $1=abc^{-1}a^{-1}\implies bc^{-1}=1$) uses the fact that the groups presentations define are invariant under cyclic reduction of the relators. – Shaun Apr 7 at 13:14
• I was just using the fact $G/[G,G]$ is abelian, but you are correct this also works (and works in $G$ so we needn't consider the abelianisation at all). – Robert Chamberlain Apr 7 at 20:17

If $$abc^{-1}a^{-1}=e$$ then $$bc^{-1}=e$$, so $$b=c$$. But then we have that $$c^3=e$$, hence the groups are not isomorphic (the free group on two letters does not contain a non-trivial torsion sub group).

Remark (as mentioned in the comments below): The fact that $$b=c$$ means that that the group $$G$$ has the presentation $$G=$$. It is clear that $$b\neq e$$, since, e.g., $$G$$ has $$S_3$$ as a quotient group. Thus $$G$$ contains an element of order $$3$$. As the Free group on two letters does not contain any non-trivial element of finite order, it is not isomorphic to $$G$$.

• There is still work to be done here. For the conclusion as you state it, you need to show that $c\neq e$ (which is not obvious). You could of course change your conclusion to something like "if $c\neq e$ then $G$ contains torsion, while if $c=e$ then $G\cong\mathbb{Z}$". In both cases, $G\not\cong F(a, b)$. – user1729 Apr 15 at 9:58
• @user1729 The group $G=<a,b\,|\,b^3>$ is not $\mathbb Z$. It has $S_3$ as a quotient, for instance. – lulu Apr 15 at 10:01
• Of course, but you have not explicitly said that $G$ is this group, just that $b=c$ and $c^3=1$. (My point is that this data is enough, when combined with the fact that $G$ is generated by $a$, $b$ and $c$, but you need to alter the conclusion. Alternatively, you could of course state that $G\cong \mathbb{Z\ast Z_3}$. But that seems like overkill.) – user1729 Apr 15 at 10:04
• @user1729 Fair enough. I'll edit accordingly. – lulu Apr 15 at 10:09

The notation $$H := \langle a, b \rangle$$ means the subgroup of $$G$$ generated by the elements $$a$$ and $$b$$. It does not mean the free group on two generators (which might be written $$F(a, b)$$ or $$\langle a, b\mid \rangle$$ or $$\langle a, b\mid - \rangle$$ or $$\langle a, b\mid \emptyset \rangle$$). Therefore, I disagree with the other two answers, and also with the working in the question.

In fact, $$G=H=\langle a, b\rangle$$. This is because the generator $$c$$ can be written in terms of the generators $$a$$ and $$b$$ (in particular, the relator $$bcb$$ implies $$c=b^{-2}$$).

If you do actually mean that $$H=F(a, b)$$ then one way of seeing this which is quite different from the other two ways is to eliminating the generator $$c$$ and obtain the presentation: $$G = \langle a, b \mid ab^3a^{-1} \rangle$$ This word is reduced, and hence $$G$$ is a proper factor group of $$F(a, b)$$. As $$F(a, b)$$ is Hopfian*, it follows that $$G\not\cong F(a, b)$$.

This method extends to other groups. For example, it proves that the group $$K=\langle a, b\mid [a^2, b^2]\rangle$$ is not free. However, the methods in the other two answers do not work for $$K$$: it is torsion free, so we cannot look at the elements of finite order, and its abelinisation is $$\mathbb{Z\times Z}$$ (the same as $$F(a, b)$$).

*A group $$K$$ is Hopfian if $$K/N\cong K\Rightarrow N=1$$. You can find a proof that free groups are Hopfian here, on Math.SE or here, on Groupprops.

• I could also not understand why people were taking $H := \langle a,b \rangle$ to mean the free group on $a$ and $b$ - why should it mean that? You could write $\langle a,b \mid \rangle$ to denote the free group but not just $\langle a,b \rangle$. – Derek Holt Apr 15 at 12:25
• I did mean $F(a, b)$. I double checked, and that is the notation my teacher uses for the free group generated by $a$ and $b$. Good to know it is not standard notation. – Pel de Pinda Apr 15 at 15:22