= Solution
A <group homomorphism> satisfies $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$; a bijective one is a <group isomorphism>. Its <kernel of a group homomorphism> is $K=\{g:\varphi(g)=1_H\}$. It is a <subgroup>, and for $k\in K$, $g\in G$,
$$
\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=1_H.
$$
Thus $gKg^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a <normal subgroup>.
Write $B_n=\begin{pmatrix}1&n\\0&1\end{pmatrix}$. Direct <matrix multiplication> gives $B_nB_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the <integer shear group>, a <subgroup> of $GL(2,\mathbb R)$. It is an <infinite cyclic group>, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.
The map $n\mapsto B_n$ is a bijective <group homomorphism> from the additive <group of integers>, proving the required <group isomorphism>. \b[There are proper subgroups]: for example $\{B_{2n}:n\in\mathbb Z\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in\mathbb Z\}$ for an integer $m\geq0$, by the classification of <subgroups of an infinite cyclic group>; $m=0$ gives the trivial subgroup and $m=1$ the whole group.
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