Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 5E / Solution

For a normal subgroup $H⊴G$, define multiplication of left cosets byTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product is $g{h}_{1}k{h}_2=gk(k^{-1}{h}_{1}k)h_2$, and normality puts $(k^{-1}{h}_{1}k)h_2$ in $H$. Thus the product coset is independent of the representatives. Associativity follows from multiplication in $G$, the identity is $H$, and the inverse of $gH$ is $g^{-1}H$. These operations form the quotient group $G/H$.

For the unheaded finite-index conclusion, let $G$ act on the $n$ left cosets of its proper subgroup $H$ by $g\cdot (xH)=gxH$. This is well-defined even if $H$ is not normal: replacing $x$ by $xh$ leaves $gxH$ unchanged. The maps are bijections with inverses supplied by $g^{-1}$, and their compositions give a group homomorphism $\phi :G\to S_n$. Its kernel of a group homomorphism $N$ is normal, by the calculation in part (ii).

The action is not trivial, because some $g\notin H$ sends the coset $H$ to $gH\ne H$; hence $N\ne G$. If $N$ were trivial, $\phi$ would be injective, implying $|G|\le |S_n|=n!$. Under the given strict inequality this is impossible. Thus $N$ is a nontrivial proper normal subgroup, andMore precisely, $N={\bigcap }_{x\in G}xH{x}^{-1}$ is the subgroup core: $g$ fixes every coset if and only if $x^{-1}gx\in H$ for every $x$. This identifies the normal subgroup produced by the coset action.