= Solution
A <right coset> of $H$ in $G$ is $Hg=\{hg:h\in H\}$. Right multiplication by $g$ is a <bijection> $H\to Hg$, with inverse $x\mapsto xg^{-1}$, so every right coset has $|H|$ elements. The analogous map $h\mapsto gh$ proves the same assertion for <left cosets>.
Write the displayed matrices as $M(a,b,c)$. Direct multiplication gives
$$
M(a,b,c)M(A,B,C)=M(a+A,b+B+aC,c+C).
$$
All new coordinates are integers, so the set is closed. Matrix multiplication is associative, the identity is $M(0,0,0)$, and
$$
\boxed{M(a,b,c)^{-1}=M(-a,ac-b,-c),}
$$
which also has integer coordinates. Thus it is the <Integer Heisenberg group>.
The subset $H=\{M(0,0,t):t\in\mathbb Z\}$ contains the identity, and multiplication and inversion add or negate $t$, so it is a <subgroup>. Its <right cosets> are
$$
\boxed{HM(a,b,c)=\{M(a,b,c+t):t\in\mathbb Z\}=\{M(a,b,s):s\in\mathbb Z\}.}
$$
They are exactly the sets with fixed integer coordinates $(a,b)$, as described by <cosets of a coordinate subgroup of the integer Heisenberg group>. The factor order $HM$ is essential; multiplying by $H$ on the other side would also change the upper-right coordinate by $at$.
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