# Is there a foolproof method to calculate the inverse of a group homomorphism?

For example, take the group $\langle a,b\mid a^{-1}b^2ab^{-3}\rangle$, and the homomorphism given induced by the map $a \rightarrow a$, $b \rightarrow b^2$. Is there a method that will let you calculate the inverse of a homomorphism of a finitely presented group?

• Do you mean the inverse of an "isomorphism" ? – Bumblebee May 30 '18 at 19:33
• Yes indeed, I only want to calculate the inverse in cases when it exists. So I suppose it would be useful to know the quickest method to check if we're working with an isomorphism or homomorphism as well – Daven May 30 '18 at 19:37
• Your example is not an isomorphism so it has no inverse. This is a standard example of a non-Hopfian group, due to Baumslag and Solitar. – Derek Holt May 31 '18 at 7:53

Well, there's a simple brute-force algorithm that always works. Let us suppose $f:\langle A\mid R\rangle\to\langle B\mid S\rangle$ is an isomorphism and we have an expression for $f(a)$ as a word in elements of $B$ for each $a\in A$. To find the inverse, take every single word $w$ in $A$ and attempt to use the relations in $S$ to reduce $f(w)$ to some element of $B$. Dovetailing these computations over all words $w$, you will eventually find for each $b\in B$ some $w_b$ such that $f(w_b)=b$. Then, the inverse is given by $f(b)=w_b$.
The Baumslag-Solitar group $\operatorname{BS}(2, 3)=\langle a, b\mid a^{-1}b^2ab^{-3}\rangle$ is not Hopfian. To see this, consider the map you give in your question, $\phi: a\mapsto a$, $b\mapsto b^2$. Then $\phi(ba^{-1}bab^{-1}a^{-1}b^{-1}a)=1$ (this is easy to see) but $ba^{-1}bab^{-1}a^{-1}b^{-1}a\neq1$ (this is harder to see - it follows from Britton's Lemma for HNN-extensions).