I'm trying to use Tietze transformations tranforms it to the group $\langle a, b, c \rangle$ - is this the right thing to do?
2026-03-25 20:42:26.1774471346
Show that the group $\langle a, b, c | ab = bac \rangle$ is free
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It can't be the free group on $3$ generators, since $c$ is completely determined in terms of $a$ and $b$: $c=a^{-1}b^{-1}ab$. Since you mention Tietze transformations, you can have a look at this one, which will help you see how to conclude.