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Past exam of the mathematics course of the University of Oxford

2014

Prelims

Mathematics I

{c}

The <group action> law gives
$$
\phi_g\circ\phi_{g^{-1}}=\phi_e=\operatorname{id}_S,
\qquad
\phi_{g^{-1}}\circ\phi_g=\operatorname{id}_S.
$$
Thus $\phi_g$ is a <bijection> with inverse $\phi_{g^{-1}}$, hence a <permutation> of $S$. Also, for every $s$,
$$
\phi_{gh}(s)=\phi(g,\phi(h,s))=(\phi_g\circ\phi_h)(s).
$$
Therefore the assignment $\Phi(g)=\phi_g$ takes values in the <symmetric group> $\operatorname{Sym}(S)$ and satisfies
$$
\boxed{\Phi(gh)=\Phi(g)\Phi(h).}
$$
It is a <group homomorphism>, with the group product on the right interpreted as composition of <permutations>.

Solution

Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 7 / a / ii / Solution

Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 7 / a / ii / Solution

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