= Solution
A <group> is a nonempty <set> with a closed <binary operation> that is <associative>, has an <identity element>, and gives every element an <inverse element>. A <subgroup> is a subset that is itself a <group> under the inherited operation. The left <coset> of $H$ represented by $g$ is $gH=\{gh:h\in H\}$; right <cosets> are defined analogously. Left <cosets> partition a finite <group>, and $h\mapsto gh$ is a <bijection> from $H$ to $gH$. Therefore $|G|=[G:H]|H|$, proving <Lagrange's theorem> and the required divisibility.
For the given <matrices>, put $a=A$ and $c=AB=iI$. Since $B=iA$, the generators commute; $a^2=I$, $c^4=I$, and the generated elements are
$$
G=\{I,c,c^2,c^3,a,ca,c^2a,c^3a\},\qquad B=ca.
$$
They are distinct, and $(c^ra^s)(c^{r'}a^{s'})=c^{r+r'}a^{s+s'}$, with exponents reduced modulo four and two. The <Cayley table> is
$$
\begin{array}{c|cccccccc}
&I&c&c^2&c^3&a&ca&c^2a&c^3a\\\hline
I&I&c&c^2&c^3&a&ca&c^2a&c^3a\\
c&c&c^2&c^3&I&ca&c^2a&c^3a&a\\
c^2&c^2&c^3&I&c&c^2a&c^3a&a&ca\\
c^3&c^3&I&c&c^2&c^3a&a&ca&c^2a\\
a&a&ca&c^2a&c^3a&I&c&c^2&c^3\\
ca&ca&c^2a&c^3a&a&c&c^2&c^3&I\\
c^2a&c^2a&c^3a&a&ca&c^2&c^3&I&c\\
c^3a&c^3a&a&ca&c^2a&c^3&I&c&c^2\\
\end{array}
$$
The table proves closure and identifies $I$ as the identity. The <inverse element> of $c^ra^s$ is $c^{-r}a^s$, and <associativity> follows from <matrix multiplication>. Thus this is the <direct product of cyclic groups of orders four and two>, an <abelian group> of order eight.
Its <subgroups> are exactly
$$
\{I\},\quad\{I,c^2\},\quad\{I,a\},\quad\{I,c^2a\},
$$
$$
\{I,c,c^2,c^3\},\quad\{I,ca,c^2,c^3a\},\quad\{I,c^2,a,c^2a\},\quad G.
$$
To see the list is complete, <Lagrange's theorem> leaves orders one, two, four and eight. There are exactly three elements of order two, giving the three order-two <subgroups>. An order-four <subgroup> containing an element of order four must be one of the two <cyclic groups> displayed; without such an element it must be the four-element set of elements whose squares are $I$. This is the enumeration in <subgroups of C4 times C2>.
For example choose $H=\{I,c,c^2,c^3\}$. A complete set of <cosets> is $\boxed{H,\qquad aH=\{a,ca,c^2a,c^3a\}}$. Since $G$ is <Abelian>, the left and right <cosets> coincide.
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