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I am searching for an example of an infinite soluble non cyclic group.

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    I am suggesting the infinite dihedral group, i.e. the group of plane motions generated by reflections with respect to two parallel lines i.e. the group of matrices of the form $$\pmatrix{\pm1&x\cr0&1\cr}.$$ – 2017-02-17

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Any non-cyclic abelian group will work. For example, take the direct product of the infinite cyclic group with any finite abelian group. Or consider the real numbers or the rational numbers or the complex numbers or... etc.

For a non-abelian example, take the direct product of your favourite infinite abelian group $A$ (e.g. the infinite cyclic group) with any finite soluble group $S$. (You should prove that the resulting group $G=A\times S$ is indeed soluble.)

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All nilpotent groups (and thus abelian) are soluble, but here's one that should feel familiar.

Take your favorite commutative ring $R$ (with identity), and construct the group of $n\times n$ upper triangular matrices, with units on the diagonal. Each matrix is therefore invertible since its determinant is a unit in $R$. You can make this example nilpotent by only having $1$s on the diagonal instead of the set of units.

Here's the case when $R=\mathbb{Z}$ and $n=3$: $$G = \left\{ \begin{bmatrix} \pm 1 & a & c \\ & \pm 1 & b \\ & & \pm 1\end{bmatrix} \;\middle|\; a,b,c\in \mathbb{Z} \right\}.$$

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    His first line is fine, it's your first line that's wrong, as your own example demonstrates! – 2017-02-17
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    @verret Gah! I've deleted the comment. – 2017-02-17