If $G_{1}$ is $\langle a,b \mid a^2 = b^2 \rangle$ and $G_{2}$ is $\langle p,q \mid pqp^{-1} = q^{-1} \rangle$, find an isomorphism $\phi : G_{1} \rightarrow G_{2}$.

I tried the obvious by letting $a$ go to $p$ and $b$ go to $q$, but that did not work.

  • $\begingroup$ The fisrt is isomorphic the second is not :) $\endgroup$ – Norbert Apr 21 '13 at 20:15

This appears to be a group theory question; I assume that $G_1, G_2$ are free groups, with the additional relations specified. Note that $(pq)^2=pqpq=pq(q^{-1}p)=p^2$.

  • $\begingroup$ Since I want to show that $\phi(ab) = \phi(a)\phi(b)$, what would my $\phi(a)$ and my $\phi(b)$ be exactly? $\endgroup$ – Richard Carpenter Apr 21 '13 at 19:40
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    $\begingroup$ I'd try $\phi(a)=p, \phi(b)=pq$. $\endgroup$ – vadim123 Apr 21 '13 at 19:42
  • $\begingroup$ See also wikipedia on Tietze transformations $\endgroup$ – Myself Apr 21 '13 at 20:36

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