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 $$<a,b,c : a^{p^{2}}=b^p=c^p=e, ab=ba^{1+p},ac=cab,bc=cb>$$ I was trying to find the embedding of the above group into a symmetric group of order $p^4$, which exists by Cayley's Theorem. Is it possible to explicitly find the map ? Kindly see it.
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Using the relations, you can represent every element uniquely as $c^k b^l a^m$ with $0\leq k < p$, $0\leq l < p$, $0\leq m< p^2$. Now you can work out how left multiplication with $a, b, c$ acts on the set of such representatives by again using the relations, this is easy to do explicitly. The corresponding action on the set of tuples $(k, l, m) $ as above is the desired embedding.
answered Aug 13, 2020 at 22:39
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$\begingroup$ Actually I have some problem in understanding your answer. For example, the left action of $a$ on $c^kb^la^m$ is given by $$a.\left( {{c^k}{b^l}{a^m}} \right) = {c^k}{b^k}{a^{1 + kp}}{b^l}{a^m} = {c^k}{b^{k + l}}{a^{1 + (k + l)p + m}}$$ how to get embedding from here ? $\endgroup$– HIMANSHUCommented Aug 14, 2020 at 19:40
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1$\begingroup$ Well, this tells you that $a$ goes to the permutation (on the set of triples $(k, l, m)$ with the bounds mentioned in the answer) which sends $(k, l, m) \mapsto (k, k+l, 1+(k+l)p+m)$. $\endgroup$ Commented Aug 14, 2020 at 22:04
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$\begingroup$ I get your point. Thanks, your answer is really helpful for me. $\endgroup$– HIMANSHUCommented Aug 14, 2020 at 22:39
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$\begingroup$ If I want to embed this image of $G$ in $S_{p^4}$ into $GL(p^4,p)$. Is there any way ? I mean, I know the concept of permutation matrices, but how to write it. I am unable to understand. Please see. $\endgroup$– HIMANSHUCommented Aug 15, 2020 at 20:37