= Solution
\b[Cosets and Lagrange.] A <left coset> is $gH=\{gh:h\in H\}$. The <left cosets> of a <subgroup> partition a <group>, each with $|H|$ elements. <Lagrange's theorem> states $\boxed{|G|=[G:H]|H|}$ for a finite <group>, so every <subgroup> order divides the <group order>.
\b[The 64-element group.] Work in the <commutative ring> $R=\mathbb Z/4\mathbb Z$, which has <zero divisors>. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct <matrix multiplication> gives
$$
\boxed{A(a,b,c)A(d,e,f)=A(a+d,b+e+cd,c+f).}
$$
All arithmetic is modulo four, so the set is closed. Associativity comes from <matrix multiplication>, the <identity element> is $A(0,0,0)$, and
$$
\boxed{A(a,b,c)^{-1}=A(-a,ca-b,-c).}
$$
The formula works in either multiplication order and requires no division. Hence these matrices form a <group> of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the <Heisenberg group over a commutative ring>.
\b[The abelian subgroup.] If $a=c$, write $h(a,b)=A(a,b,a)$. Then
$$
h(a,b)h(d,e)=h(a+d,b+e+ad)=h(d,e)h(a,b).
$$
The identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an <abelian subgroup>. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with <Lagrange's theorem>. This is the <abelian plane subgroup of the Heisenberg group modulo four>.
More explicitly, $A(a,b,c)\mapsto a-c$ is a <surjective group homomorphism> to the additive group of $R$, with kernel $H$. The four <left cosets> are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a <normal subgroup>.
\b[Matrices squaring to the identity.] The product formula gives
$$
A(a,b,c)^2=A(2a,2b+ca,2c).
$$
The first and third coordinates vanish only when $a,c\in\{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in\{0,2\}$. Therefore the complete set is
$$
\boxed{K=\{A(a,b,c):a,b,c\in\{0,2\}\},\qquad |K|=8.}
$$
It is a <subgroup>: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the <direct product of groups> $C_2\times C_2\times C_2$. The seven nonidentity elements are <involutions>.
Finally, the last subset in the PDF is defined by $a\ne0\ne b$, with no restriction on $c$. It has $3\cdot3\cdot4=36$ elements, which cannot be a <subgroup> because $36$ does not divide $64$. It also excludes the <identity element>. Thus
$$
\boxed{\{A(a,b,c):a\ne0,\ b\ne0\}\text{ is not a subgroup}.}
$$
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