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In the beginning steps of Strong Cayley theorem, I got stuck at a point.

$H\leqslant G,\,\Sigma=\{Hx\mid x\in G\}$, we define \begin{align*} P~:~G&\longrightarrow S(\Sigma),\\g&\longmapsto Hxg. \end{align*} The permutation representation is such: \begin{gather*} P(g)=\left(\begin{array}{llll} Hx_1&Hx_2&\cdots&Hx_n\\ Hx_1g&Hx_2g&\cdots&Hx_ng \end{array}\right)=\left(\begin{array}{l} Hx_i\\ Hx_ig \end{array}\right). \end{gather*} It's easy to show $P$ is a homomorphism: \begin{gather*} P(a)\circ P(b)=\left(\begin{array}{l} Hx_i\\ Hx_ia \end{array}\right)\circ\left(\begin{array}{l} Hx_i\\ Hx_ib \end{array}\right)~,\forall a,b\in G. \end{gather*} This shows that $P$ changes $Hx_i$ into $Hx_ib$ firstly, then $Hx_i$ into $Hx_ia$: \begin{gather*} P(a)\circ P(b)=P(a)\circ \left(\begin{array}{l} Hx_i\\ Hx_ib \end{array}\right) =\left(\begin{array}{l} Hx_i\\ (Hx_ia)b \end{array}\right)=\left(\begin{array}{l} Hx_i\\ Hx_i(ab) \end{array}\right)=P(ab). \end{gather*}

However, when I employ the right induced representation : We set $\Omega=\{xH\mid x\in G\},$ \begin{align*} O~:~G&\longrightarrow S(\Omega),\\g&\longmapsto gxH, \end{align*} I can't find the way to show that $O$ is still a homomorphism.

Is there anything that I miss or understood? Any help is sincerely appreciated.

PS: When we apply the same logic in the map O as P, we’ll get: $$(O(a)O(b))(xH)= O(a)(O(b)(xH)) =(Hxb)a=Hx(ba)=O(ba),$$which is ridiculous considering the acknowledged fact that $O(ab)=O(a)\circ O(b)$. Why?

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So the map $O$ sends $g\in G$ to the permutation on $\Omega$ defined by $O(g)(xH)=gxH$, for $xH\in\Omega$. Then for $g,h\in G$, and arbitrary $xH\in\Omega$, $$ O(gh)(xH)=ghxH=O(g)(hxH)=O(g)O(h)(xH), $$ so that $O(gh)=O(g)O(h)$ as permutations on $\Omega$.

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  • $\begingroup$ When we apply the same logic in the map P, we’ll get: $P(ab)(Hx)=(Hxa)b=P(b)\circ (Hxa)=P(b)\circ P(a)$, considering that $P(ab)=P(a)\circ P(b)$, it will be ridiculous. Why? $\endgroup$ – user517681 Feb 14 '18 at 7:31
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    $\begingroup$ @Benny The second equality in the last line of your quoted block is incorrect, since $P(a)(Hx_ib)=Hx_iba)$. Right multiplication in this case is an antihomomorphism, $P(ab)=P(b)P(a)$. You can make a homomorphism by defining instead $P(a)(Hx)=Hxa^{-1}$. $\endgroup$ – BW. Feb 14 '18 at 7:39
  • $\begingroup$ I should add there's not really any content to the discrepancies between what is a homomorphism and antihomomorphism here, it's mostly due to the convention of writing function composition on the left. $\endgroup$ – BW. Feb 14 '18 at 7:47
  • $\begingroup$ Thank you. But I do not quite understand what you mean by “convention of writing function composition”, as which was the same answer from teacher. $\endgroup$ – user517681 Feb 14 '18 at 8:27
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    $\begingroup$ That is surprising to me! Thanks! $\endgroup$ – user517681 Feb 14 '18 at 8:39

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