= Solution
Denote a matrix by $A(x,y,z)$, with $y$ the upper-right entry. Direct <matrix multiplication> gives
$$
\boxed{A(x,y,z)A(x',y',z')=A(x+x',\ y+y'+xz',\ z+z').}
$$
This proves closure. Associativity is inherited from matrix multiplication. The identity is $A(0,0,0)$, and the inverse is
$$
\boxed{A(x,y,z)^{-1}=A(-x,xz-y,-z).}
$$
Both products with this matrix are the identity. Thus these real matrices form the <Heisenberg group>, in the coordinate convention with central coordinate $y$.
Let $N=\{A(0,b,0):b\in\mathbb R\}$. Its multiplication adds $b$, its identity is $b=0$ and its inverse negates $b$, so it is a <subgroup>. Moreover
$$
A(x,y,z)A(0,b,0)=A(x,y+b,z)=A(0,b,0)A(x,y,z).
$$
Thus every element of $N$ is central, and $gNg^{-1}=N$. Therefore \b[$N$ is a normal subgroup]. In fact it is the full <center of a group>, as described by the <center quotient of the real Heisenberg group>.
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