Exercise :
Show that the group $\mathbb Z_{45}$ has exactly one subgroup of order $3$ and construct it.
Discussion :
First of all, $\mathbb Z_{45}$ has subgroups of order $d$, where $d$ is a divisor of the order of the given group, $\mathbb Z_{45} = 45$. Truly then, if $d=3$, it divides $45$, as it is $45/3 = 15$, so $\mathbb Z_{45}$ has a subgroup of order $3$.
The uniqueness and construction part is where I get confused though.
Let's suppose that $\mathbb Z_{45}$ has two subgroups $H,G$ of order $3$. Then, since $\mathbb Z_{45}$ is abelian, it will be $HG = GH$ and $HG$ would be a subgroup of $\mathbb Z_{45}$. But $|HG| = 3^2/|H \cap G| = 3\cdot d$ where $d=3,5,9,15$. But this contradicts to Lagrange's Theorem, so the subgroup of order $3$ should be unique.
Is this approach for the uniqueness correct ? For the construction part, I really do not know how to proceed (I generally have a difficulty of constructing groups in such cases and I would really appreciate a thorough elaboration or some tips on how to make my life easier on this).
Take $H=\{0,15,30\}$. Then $H$ is a subgroup of $\mathbb{Z}_{45}$ of order $3$.
Does $\mathbb{Z}_{45}$ have other elements of order $3$ besides $15$ and $30$? No, because $45\mid3n\iff15\mid n$. Since hevery subgroup of $\mathbb{Z}_{45}$ must have $2$ elements of order $3$, this shows that $H$ is the only such subgroup.
Concerning your approach, I don't see where is it that you get a contradiction with Lagrange's theorem.