= Solution
A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomes
$$
A(t,b,t)A(s,d,s)=A(t+s,\ b+d+ts,\ t+s).
$$
It stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t)^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an <Abelian subgroup>, namely the <Abelian plane subgroup of a finite Heisenberg group> in this coordinate convention.
There are five choices each for $t,b$, so
$$
\boxed{|H|=25,\qquad [G:H]=\frac{125}{25}=5.}
$$
This verifies <Lagrange's theorem>: the subgroup order divides the group order, and the number of distinct left <cosets> is five.
Explicitly, the map $\varphi(A(a,b,c))=a-c$ is a surjective <group homomorphism> to the additive group of $\mathbb F_5$, by the product formula. Its <kernel of a group homomorphism> is $H$. Thus $H$ is even a <normal subgroup>; the five left <cosets> are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.
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