Minimum number of elements in group Suppose we have $G=\langle(123),(456),(14)(25)(36)\rangle$ a subgroup of $S_6$
Am I correct in saying that the minimum number of elements in $G$ equals the product of the orders of the elements of the generating set?
$|(123)|=3$, $|(456)|=3$ and $|(14)(25)(36)|=2$, so we have a minimum of $3\cdot 3 \cdot 2 = 18$ elements generated. That means $18 \leq |G|$.
Is is possible in general to multiply the orders of the different elements in the generating set of a group that is not abelian and say that is the minimum number of elements? 
 A: No. Consider the group $S_4$. It is generated by the six transpositions $(12)$, $(13)$, $(14)$, $(23)$, $(24)$ and $(34)$. The order of the group is $4!$ - much less that $2^6=64$

I was more than a bit unhappy with the above example. Its success exploits a kind of sloppiness in the formulation of the question. Namely the fact that I used a non-minimal generating set. To wit, it suffices to use only the 2-cycles $(12)$, $(13)$ and $(14)$ to get all of $S_4$, and the OPs guess is valid for that set of generators. Therefore the above counterexample is lacking in intellectual honesty.
My suggested remedy:


*

*Let's add the constraint that no proper subset of the chosen set of generators will generate the same group.

*Consider the group of eight unit quaternions
$$
Q_8=\{\pm1,\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k}\}
$$
defined by the relations $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-1=\mathbf{ijk}$ (together with the obvious effect of multiplication by $-1$).
We see that $Q_8$ is not cyclic, so at least two generators are needed. However,
$Q_8$ is generated by $\mathbf{i}$ and $\mathbf{j}$. Both of those are of order four, and $4\cdot4=16>|Q_8|$. Therefore this is an honest counterexample.

*The counterexample of the previous bullet works even if we insist on using a set of generators such that the product of their orders is minimized. In the case of $Q_8$ any generating set must include at least two elements of order four.

