# A Question Raised about Strong Cayley Theorem

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?

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 $\Omega$.
• 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? – user517681 Feb 14 '18 at 7:31
• @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 $P(a)(Hx)=Hxa^{-1}$. – BW. Feb 14 '18 at 7:39