Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 7D (Groups)

Lagrange's theorem says that, for a finite group $G$ and subgroup $H$,Here $[G:H]$ is the number of left cosets. Every $g\in G$ lies in $gH$. If $gH$ and $g^{\prime }H$ meet, write $gh=g^{\prime }{h}^{\prime }$; then $g^{\prime }=gh(h^{\prime }{)}^{-1}$, so $g^{\prime }H=gH$. Hence the left cosets partition $G$. The bijection $h\mapsto gh$ makes each coset have $|H|$ elements. Adding their cardinalities proves Lagrange's theorem.

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus the double cosets partition the group.

Unlike single cosets, double cosets can have different cardinalities. For finite $G$, decompose $HgK$ into left cosets of $K$, or apply the orbit-stabilizer theorem to the action of $H$ on $gK$. Its stabiliser subgroup is $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.

False. Let $G=S_3$ and $H=K=⟨(12)⟩$. For $g=(13)$, direct multiplication gives the double cosetIt has four elements, and $4\nmid 6=|S_3|$. This is a concrete example of double-coset sizes not dividing group order; Lagrange's theorem applies to subgroups, not to arbitrary double cosets.

False. Use the same subgroups of $S_3$. The double coset $HeH=H$ has two elements, while $H(13)H$ has four. Thus even a fixed pair of subgroups can give different double-coset sizes. Their disjoint union is all six elements of $S_3$.

False. Realize the symmetries of a square as a dihedral group of order eight, generated by $r=(1234)$ and $s=(24)$, with $r^4=s^2=e$ and $srs=r^{-1}$. Take $H=K=⟨s⟩$. Its double cosets are preciselyThese are disjoint and exhaust the eight symmetries. There are three double cosets, and $3\nmid 8$. This proves that double-coset counts need not divide group order. Specifying the order-eight realization avoids competing conventions for the notation of a dihedral group.