= 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
|| $\cdot$
|| $I$
|| $A$
|| $B$
|| $C$
| $I$
| $I$
| $A$
| $B$
| $C$
| $A$
| $A$
| $I$
| $C$
| $B$
| $B$
| $B$
| $C$
| $I$
| $A$
| $C$
| $C$
| $B$
| $A$
| $I$
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>.
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