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In the (xi) group of the classification of groups of order $p^4$ given by W.Burnside in his book," Theory of Groups Of Finite Order". The group ($\mathbb{Z_{p^{2}}}\rtimes \mathbb{Z_{p^{}}}) \rtimes_{\phi}\mathbb{Z_{p^{}}} $, have presentation $$<a,b,c : a^{p^{2}}=b^p=c^p=e, ab=ba^{1+p},ac=cab,bc=cb>$$ From the above relations, I can produce following relation $$a^ib^j=b^ja^{{(1+p)}^ji},$$ Now I am trying to obtain similar relation for the generators $a$ and $c$. (Here $i$ and $j$ are natural numbers). Kindly help me out as I am not able to proceed further to write $a^ic^j= ? ?$ in simple manner.

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$$a^ic^j = c^j(ab^j)^i.$$ I expect you could use your existing formula for $a^ib^j$ to write $(ab^j)^i$ in the form $b^ka^l$, but I will leave that to you!

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  • $\begingroup$ I got my final answer as $$a^ic^j=c^jb^{ij}a^{i+\frac{ji(i+1)}{2}p.}$$ $\endgroup$ – Setia H Jun 27 at 11:16
  • $\begingroup$ Yes I checked a few cases, and that seems to be correct! $\endgroup$ – Derek Holt Jun 27 at 12:12

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