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I am trying to understand a proof given of an isomorphism between an infinite and finite presentation of Thompson's group F in the following paper by Cannon, Floyd and Parry.

I'm referring to Theorem 3.1 on page 7.

So, I understand how to construct the surjective homomorphism (I'll just call it $\phi$) mentioned at the beginning. It's the steps after this that I don't quite understand. How does:

$(i)$ Showing that the defining relations of $F_1$ are contained in the kernel of $\phi$

$(ii)$ Showing that there exists a homomorphism the other way (from $F_2$ to $F_1$)

Prove the result?

To show $\phi$ is injective surely you have to show that the defining relations of $F_1$ are equal to ker($\phi$)? (since then $\phi$ has trivial kernel)

Also, since $\phi$ is defined by $\phi(A) = X_0$ and $\phi(B) = X_1$, does showing there exists a homomorphism the other way that maps $X_0$ to $A$ and $X_1$ to $B$ not prove that $\phi$ has an inverse, and so must be an isomorphism anyway? If so, why bother to show $(i)$?

Thanks in advance for any help

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Maybe some people are put off by the paper on Thompson groups? Anyway, the question isn't really related to this. All I'm really asking is how does the following imply an isomorphism between group presentations $F_1$ and $F_2$: $(i)$ There exists a surjective homomorphism $\phi: F_1 \rightarrow F_2$. which maps $A$ to $X_0$ and $B$ to $X_1$ $(ii)$ the defining relations of $F_1$ are contained in the kernel of $\phi$ $(iii)$ There exists a homomorphism from $F_2$ to $F_1$ which maps $X_0$ to $A$ and $X_1$ to $B$ – Kotov Feb 12 '14 at 17:32
The map $\phi$ and the homomorphism the other way are mutually inverse maps, and so both are bijections. – Derek Holt Feb 12 '14 at 17:33
Thanks for your response. Then why is it necessary in the proof to show that the defining relations of $F_1$ are contained in ker($\phi$)? – Kotov Feb 12 '14 at 17:36
The original homomorphism is from the free group. You have to prove that the relations of $F_1$ are in the kernel in order to get an induced homomorphism from $F_1$ to $F_2$. It's the same for the inverse map. – Derek Holt Feb 12 '14 at 17:41
Maybe I have figured it out: Perhaps the above condition $(i)$ is required for $\phi$ to be a homomorphism since otherwise $\phi$ does not satisfy the property $\phi(1) = 1$? – Kotov Feb 12 '14 at 17:42
up vote 0 down vote accepted

Let me explain the general framework of the proof.

Let $G_1 = \langle X_1 \mid R_1 \rangle$ be a group presentation, where $R_1$ is a set of relators. So $G_1 = F_1/\langle R_1^{F_1} \rangle$, where $F_1$ is the free group on $X_1$.

Now if $H$ is any group, and $\phi:X_1 \to H$ is any map, then $\phi$ extends uniquely to a homomorphism $F_1 \to H$ and, if $R_1 \le \ker \phi$, then $\phi$ induces a homomorphism $\bar{\phi}_1:G_1 \to H$.

In this proof, we have another group presentation $G_2 = \langle X_2 \mid R_2 \rangle$, and the homomorphism $\bar{\phi}_1:G_1 \to G_2$ is defined as above.

Then the equivalent thing is done in the other direction, and a homomorphism $\bar{\phi}_2:G_2 \to G_1$ is defined.

It is straightforward to check in this example that $\bar{\phi}_2\bar{\phi}_1(x)=x$ for all $x \in X_1$ and $\bar{\phi}_1\bar{\phi}_2(x)=x$ for all $x \in X_2$. This proves that $\bar{\phi}_2\bar{\phi}_1:G_1 \to G_1$ and $\bar{\phi}_1\bar{\phi}_2:G_2 \to G_2$ are both equal to the identity maps. So $\bar{\phi}_1$ and $\bar{\phi}_2$ are mutually inverse maps, and hence both are isomorphisms.

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