= Solution
For a <normal subgroup> $H\trianglelefteq G$, define multiplication of left <cosets> by
$$
\boxed{(gH)(kH)=(gk)H.}
$$
To check that this is well-defined, replace $g,k$ by $gh_1,kh_2$ with $h_1,h_2\in H$. Their product is $gh_1kh_2=gk(k^{-1}h_1k)h_2$, and normality puts $(k^{-1}h_1k)h_2$ 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> $\varphi:G\to S_n$. 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, $\varphi$ would be <injective>, implying $|G|\leq|S_n|=n!$. Under the given strict inequality this is impossible. Thus $N$ is a nontrivial proper <normal subgroup>, and
$$
\boxed{|G|>n!\ \Longrightarrow\ G\text{ is not simple}.}
$$
More precisely, $N=\bigcap_{x\in G}xHx^{-1}$ 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>.
Back to article page
