$5$. Consider the set $G = (1, 3, 5, 7, 9, 11, 13, 15)$
(a) Show that this set is closed under the binary operation x, y → xy (mod 16)
and write down the corresponding multiplication table. [8]
(b) Show that G is a group under this operation and find all its cyclic subgroups. [6]
(c) Is G isomorphic to Z8? Justify your answer. [6]
(d) State Cayley’s theorem and find a subgroup in S8 which is isomorphic to G...
I have completed parts a, b and c but d is giving me some trouble.
I can see that the subgroup will be (3,9,11,1) based on my answer to part c, but how that justifies 10 marks in part d is beyond me. I'm not sure where the problem lies.
Cayley's Theorem: Any finite group of order n is isomorphic to a subgroup of symmetric group $S_n$
In looking at the Cayley's theorem proof, I cant find the way to apply it to my specific problem, so if someone could point me in the right direction that would be much appreciated.
Cheers
As for d), you have got to literally stuck to what the theorem states, namely: $$G\stackrel{\varphi}{\cong} K :=\{1\cdot\_\space,3\cdot\_\space,5\cdot\_\space,7\cdot\_\space,9\cdot\_\space,11\cdot\_\space,13\cdot\_\space,15\cdot\_\space\}\le S_G \tag1$$ where, for all $i\in G$, "$i\cdot\_$" is the bijection on $G$ defined by $j\mapsto i\cdot j\pmod{16}$.
And $K$ can be "isomorphically transferred" into $S_8$ via $\psi_f\colon S_G \to S_8$, defined by $\sigma\mapsto f\sigma f^{-1}$ for any bijection $f\colon G\to\{1,\dots,8\}$. For example, the choice: $$f(1_G)=1,\space f(3_G)=2,\space f(5_G)=3,\space f(7_G)=4,\space f(9_G)=5,\space f(11_G)=6,\space f(13_G)=7,\space f(15_G)=8$$ yields: \begin{alignat}{1} (\psi_f(1_G\cdot\_))(1) &= (f(1_G\cdot\_) f^{-1})(1) \\ &= f((1_G\cdot\_)(f^{-1}(1))) \\ &= f(1_G\cdot 1_G\pmod{16}) \\ &= f(1_G) \\ &= 1 \end{alignat} \begin{alignat}{1} (\psi_f(1_G\cdot\_))(2) &= (f(1_G\cdot\_) f^{-1})(2) \\ &= f((1_G\cdot\_)(f^{-1}(2))) \\ &= f(1_G\cdot 3_G\pmod{16}) \\ &= f(3_G) \\ &= 2 \end{alignat} \begin{alignat}{1} (\psi_f(1_G\cdot\_))(3) &= (f(1_G\cdot\_) f^{-1})(3) \\ &= f((1_G\cdot\_)(f^{-1}(3))) \\ &= f(1_G\cdot 5_G\pmod{16}) \\ &= f(5_G) \\ &= 3 \end{alignat} etc.. So, finally: $$\psi_f(1_G\cdot\_)=()$$ Likewise, you will find: \begin{alignat}{1} &\psi_f(3_G\cdot\_) &&=(1256)(3874) \\ &\psi_f(5_G\cdot\_) &&=(1357)(2864) \\ &\psi_f(7_G\cdot\_) &&=(14)(23)(58)(67) \\ &\psi_f(9_G\cdot\_) &&=(15)(26)(37)(48) \\ &\psi_f(11_G\cdot\_) &&=(1652)(3478) \\ &\psi_f(13_G\cdot\_) &&=(1753)(2468) \\ &\psi_f(15_G\cdot\_) &&=(18)(27)(36)(45) \\ \end{alignat} whence: $$G\stackrel{\psi_f\varphi}{\cong}\{(),(1256)(3874),(1357)(2864),(14)(23)(58)(67),(15)(26)(37)(48),(1652)(3478),(1753)(2468),(18)(27)(36)(45)\}\le S_8$$ (so, incidentally, $G\cong C_2\times C_4$). It may worth noting that $G\hookrightarrow S_8$ non-canonically ("$f$").