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An infinite cyclic group is generated by an element of infinite order and is isomorphic to the additive group of integers.

The lamplighter group is the restricted wreath product $C_2\wr \mathbb{Z}$. An element consists of a finitely supported function $f:\mathbb{Z}\to {\mathbb{F}}_2$, encoding the on/off lamps, and an integer position $n$. Let ${\tau }_{n}g(k)=g(k-n)$. The group operation isOnly finitely many lamps are on, so the configuration group is the restricted direct sum of groups ${⨁}_{k\in \mathbb{Z}}{C}_2$. The position group shifts their indices, giving the displayed semidirect product.

Let $a=({\delta }_0,0)$ switch the lamp at zero and let $t=(0,1)$ move the position by one. Then $t^{k}a{t}^{-k}=({\delta }_k,0)$ switches the lamp at $k$. Products of these switches create any finite configuration, so $a,t$ are a generating set of a group. Although every switch has order two, $t$ has infinite order; the group itself is infinite and nonabelian. Projection $(f,n)\mapsto n$ has the lamp configuration group as its group kernel.

We can use reduction instead of expanding a very large rational addition formula. Modulo $2$, both $P_1$ and $P_2$ reduce to $(0,0)$, which generates the three-element reduced group. The reduction homomorphism therefore givesThe rational point $P_2$ is not any of the three torsion points just found, so it has infinite order. In particular $R=P_1+5P_2\ne O$: otherwise multiplication by three would give $15P_2=O$.

If $R$ had integral coordinates, its affine integral representative would reduce to an affine point modulo $2$, not to the point at infinity. This contradicts its reduction to $O$. ThusIndeed, it is not even integral at $2$. This is a reduction certificate for a nonintegral elliptic point, with the crucial nonidentity condition verified by the torsion calculation.

Introduce $b=t^{-1}at$. The presentation may be written as the HNN extensionof the Baumslag-Solitar group $\mathrm{BS}(1,2)=⟨a,b\mid ba{b}^{-1}=a^2⟩$, with stable letter $t$ identifying the infinite cyclic subgroups $⟨a⟩$ and $⟨b⟩$. Britton's lemma embeds the base group in the HNN extension. In particular, $a$ has infinite order, so $G$ is infinite. It is also nonabelian because $ba{b}^{-1}=a^2\ne a$.