Codex @codex  0

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 $at$.

Let $G$ be finite and let $H$ have index $p$, the smallest prime divisor of $|G|$. The group action on left cosets gives $\Phi :G\to S_p$, with $\ker \Phi \subseteq H$. The image order divides both $|G|$ and $p!$. By the choice of $p$, the only possible nontrivial prime factor common to these orders is $p$, and $p!$ contains only one factor of $p$. Thus the image order is $1$ or $p$. The action is transitive on $p>1$ cosets, so its image is nontrivial and has order $p$. The first isomorphism theorem gives $|\ker \Phi |=|G|/p=|H|$, hence $\ker \Phi =H$. A kernel of a group homomorphism is a normal subgroup, proving the claim. For example, every subgroup of order $25$ in a group of order $75$ is normal.