Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 3 / 2D / Solution

Lagrange's theorem says that for a subgroup $H$ of a finite group $G$,To prove it, the left cosets $gH$ partition $G$. Indeed every element lies in its own coset; if $gH$ and $kH$ intersect, write $gh=k{h}^{\prime }$ and deduce $g=k{h}^{\prime }{h}^{-1}$, so $gH=kH$. Multiplication by $g$ is a bijection from $H$ to $gH$, and each coset therefore has $|H|$ elements. Summing over the $[G:H]$ distinct cosets proves the formula.

For the failure of the converse to Lagrange's theorem, take the alternating group $A_4$, of order $4!/2=12$, and the divisor $6$. Suppose it had a subgroup $H$ of order six. Its index would be two. An index-two subgroup is normal: for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, while for $g\in H$ they are $H$. Hence $A_4/H$ would be a group of order two.

Every three-cycle in $A_4$ has order three. Its image in the quotient has cube equal to the identity and, since the quotient has order two, must itself be the identity. Thus all three-cycles lie in $H$. There are eight of them: choose the omitted point in four ways and one of two cyclic orders on the other three points. Together with the identity this would give at least nine elements of $H$, contradicting $|H|=6$. Therefore $A_4$ has no subgroup of order six, although six divides its order.