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 $n$).


1 Answer 1


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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .