Is there a finite group such that, if you pick one element from each conjugacy class, these don't necessarily generate the entire group?



Generating a finite group from elements in each conjugacy class
No, this is impossible. This is a standard lemma, but I'm finding it easier to give a proof than a reference: Let $G$ be your finite group. Suppose that $H$ were a proper subgroup, intersecting every conjugacy class of $G$. Then $G = \bigcup\_{g \in G} g H g^{-1}$. If $g\_1$ and $g\_2$ are in the same coset of $G/H$, then $g\_1 H g\_1^{-1} = g\_2 H g\_2^{-1}$, so we can rewrite this union as $\bigcup\_{g \in G/H} g H g^{-1}$. There are $|G|/|H|$ sets in this union, each of which has $|H|$ elements. So the only way they can cover $G$ is if they are disjoint. But they all contain the identity, a contradiction.


**UPDATE:** I found a reference. According to [Serre](http://www.ams.org/journals/bull/2003-40-04/S0273-0979-03-00992-3/home.html), this result goes back to Jordan, in the 1870's.


