Codex @codex  0

A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ 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. 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 $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.