# Smallest subgroup generated by a subset of a group.

If$$\ G$$ is a group, and the set $$S=\{a,b\}$$ is a subset of $$\ G$$, can we say that the smallest subgroup of $$\ G$$ generated by $$\langle a,b\rangle$$ will always be either $$\langle a \rangle$$, $$\langle b\rangle$$, or in the case that $$\langle a\rangle$$ does not generate $$b$$ and $$\langle b\rangle$$ does not generate $$a$$, then $$\langle a,b\rangle$$ = $$G$$?

I'm having difficulty trying to think of a counterexample to this, particularly a finite group whose elements can be easily enumerated (e.g. $$\ U(n)$$ the multiplicative group of integers modulo $$n$$).

No. Just take the group $$C_2 \times C_2 \times C_2$$ to see this. To motivate a little the construction of an example of a group which fails to satisfy your property, think of the situation like this:
If we can find a proper subgroup $$H$$ of $$G$$ such that $$H$$ is not cyclic, $$G$$ will definitely fail to satisfy your property. Do you see why?
Then it is easy to see that the smallest non-cyclic group is $$V = C_2 \times C_2$$. How can we have $$V$$ as a proper subgroup of some other group? The most obvious way to do that is to take $$G = A \times V$$, where $$A>1$$ is any group. The example I gave is the choice $$A = C_2$$ (and is a group of smallest possible order which fails to satisfy your property).