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Write $M(a,b,c)=\left(\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}\right)$ with integer coordinates. Multiplication in this parametrization is $M(a,b,c)M(A,B,C)=M(a+A,b+B+aC,c+C)$. The subgroup $H=\{M(0,0,t):t\in \mathbb{Z}\}$ has right cosets obtained by multiplying on the left by its elements, so its right cosets are exactly the displayed sets with fixed $(a,b)$. In particular, right cosets are indexed by ${\mathbb{Z}}^2$. Left cosets differ: $M(a,b,c)H$ fixes $a$ and $b-ac$, since right multiplication changes $b$ by $at$ and $c$ by $t$. This also exhibits why $H$ is not a normal subgroup of the Integer Heisenberg group.

For a subgroup $H$ of a group $G$ and $g\in G$, its left coset is $gH=\{gh:h\in H\}$. Left multiplication $h\mapsto gh$ is a bijection from $H$ onto $gH$. Left and right cosets agree for every $g$ exactly when $H$ is a normal subgroup.

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 $x\mapsto x{g}^{-1}$, 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 givesAll new coordinates are integers, so the set is closed. Matrix multiplication is associative, the identity is $M(0,0,0)$, andwhich 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 areThey 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 $at$.