# concrete equality of group presentations

Why does this equality hold? $$\begin{array}{rl} \langle a,c| &ca^{-1}cac^{-1}aca^{-1}c^{-1}ac^{-1}a^{-1}ca^{-1}c^{-1}a,\\ &ac^{-1}aca^{-1}cac^{-1}a^{-1}ca^{-1}c^{-1}ac^{-1}a^{-1}c\rangle\\ =\langle a,c| &aca^{-1}cac^{-1}aca^{-1}c^{-1}ac^{-1}a^{-1}ca^{-1}c^{-1}\rangle\\ \end{array}$$ I'm asking this cause I don't understand the last line of this calculation:

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Leon: is this more or less what you wanted? –  t.b. Jun 9 '11 at 12:59
yes, thank you. I repaired the syntax myself, but you were quicker. thanks –  Leon Jun 9 '11 at 13:00
If I see correctly, the relation in the second group is simply obtained from the first relation in the first group by conjugating with $a$. So the question is: what is the connection between the first and the second relation in the first group. –  t.b. Jun 9 '11 at 13:26

## 1 Answer

As Theo remarked in the comments, the relation in the second group is obtained from the first relation in the first group by conjugating with $a$.

The second relation in the first group is obtained from the first relation in the first group by inverting it and then conjugating with $c^{-1}a$.

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Yes, that's it! –  t.b. Jun 9 '11 at 13:41
uh, you're right. this was non-trivial... thank you –  Leon Jun 9 '11 at 13:46