Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 8 / Solution

A left coset is $gH=\{gh:h\in H\}$. The cosets partition a finite group into equally sized sets of $|H|$ elements, so Lagrange's theorem gives $|G|=[G:H]|H|$; in particular subgroup order divides group order.

Here $A=-I$, $C=-B$ and $B^2=I$. The Cayley table, with row element multiplied on the left of the column element, is

The table proves closure, $I$ is the identity and every element is its own inverse. Associativity is inherited from matrix multiplication, so this is the Klein four-group.

Finally, if every nonidentity element has order two, every element satisfies $h^{-1}=h$, including the identity. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian.