= Solution
Perform the product and inverse formulas from part (b) in the commutative ring $\mathbb Z/4\mathbb Z$. They involve only addition, multiplication and negatives, so closure and inverses remain valid; no division by a nonunit is needed. Associativity again comes from matrix multiplication, and $A(0,0,0)$ remains the identity. The <Heisenberg group over a commutative ring> therefore has
$$
\boxed{|G|=4^3=64.}
$$
These are residues modulo four, not arithmetic in the <finite field> of four elements.
For the additional subgroup task, write an element of $H$ as $A(t,y,t)$. Its product and inverse are
$$
A(t,y,t)A(s,v,s)=A(t+s,y+v+ts,t+s),\qquad A(t,y,t)^{-1}=A(-t,t^2-y,-t).
$$
Both remain in $H$, and the product is unchanged on exchanging the factors because $ts=st$. The identity is included, so $H$ is an <Abelian subgroup>. There are four independent choices each for $t,y$, giving
$$
\boxed{|H|=16,\qquad[G:H]=\frac{64}{16}=4.}
$$
Thus there are four distinct left <cosets>. Explicitly the surjective <group homomorphism> $A(x,y,z)\mapsto x-z\pmod4$ has kernel $H$; the four cosets are the level sets of that difference, with representatives $A(d,0,0)$, $d=0,1,2,3$. This is the <Abelian plane subgroup of the Heisenberg group modulo four>.
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