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

$\begingroup$ Actually, I would guess that since $SO(3)$ has this property, it wouldn't be too hard to find a finite subgroup of $SO(3)$ which also does. $\endgroup$– Jamie VicaryCommented Jun 3, 2010 at 22:05

3$\begingroup$ I don't see why there should be a connection: it's certainly not the case for finite cyclic and finite dihedral subgroups of $SO(3)$. $\endgroup$– Victor ProtsakCommented Jun 4, 2010 at 0:05

$\begingroup$ Not sure why this question got resurrected but while it’s up here a negative answer follows from the class equation 1 = \sum_C 1/#Z(C) valid for all groups (C are the conjugacy classes and Z(C) is the centralizer of any element in the conjugacy class). When passing to such a subgroup the number of conjugacy classes goes up (here is where the hypothesis gets used), but the sizes of the centralizers go down. $\endgroup$– alpogeCommented May 16, 2019 at 19:05
5 Answers
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, this result goes back to Jordan, in the 1870's.
It is impossible. As I mentioned in the comment to Richard Stanley's answer, you are looking for a finite group $G$ with a maximal subgroup $M$ such that $M$ intersects every conjugacy class. Then $G=\cup M^g$ is the union of $M$ and its conjugates, which is wellknown to never happen.
Steve

2$\begingroup$ "Wellknown" as in "an easy exercise". $\endgroup$– Steve DCommented Jun 4, 2010 at 0:12

1$\begingroup$ Wellknown to those who know it, I suppose. What (precisely) are saying never happens? $\endgroup$ Commented Jun 4, 2010 at 0:14

6$\begingroup$ It's worth pointing out that this argument really uses finite. For example, for compact simple Lie groups every element lies in some torus. $\endgroup$ Commented Jun 4, 2010 at 0:20

1$\begingroup$ Very minor nitpick, you mean $(M1)[G:M]+1 < G$ $\endgroup$ Commented Jun 4, 2010 at 0:22

1$\begingroup$ The easiest example in infinite groups may be the group of invertible upper triangular matrices, which meets every conjugacy class of ${\rm GL}(n,k)$ when $k$ is an algebraically closed field. This doesn't require the full strength of the Jordan Normal Form theory. $\endgroup$ Commented Apr 28, 2011 at 19:32
The impossibility also follows from Jordan's lemma:
Let $G$ be a finite group which acts transitively on a set $\Omega$ with $\Omega:=n\geq 2$. Then there exists a $g\in G$ such that $\chi(g)=0$ where $\chi$ denotes the permutation character (put in simple terms this means that $g$ fixes no element of $\Omega$ ).
In fact with some additional work one can show that the proportion of elements $g\in G$ such that $\chi(g)=0$ is larger than or equal to $\frac{1}{n}$. So now let us see how Jordan's lemma implies that the answer to the OP's question is negative. So let $H$ be the group generated by $\{g_i\}$, a complete set of representatives of the conjugacy classes of $G$. Suppose that $H$ is a proper subgroup of $G$. Then we may look at the left action of $G$ on $G/H$. Since $G/H\geq 2$ and the action is transitive, it follows from Jordan's lemma that there exists a $x\in G$ such that for all $i$, $x g_i H\neq g_i H$. In other words, for each $g_i$ one has that $g_i^{1}x g_i\notin H$ which in turn implies that for all $g\in G$ one has that $g^{1}xg\notin H$; and therefore the conjugacy class of $x$ does not intersect $H$ which is absurd.
Note also that one gets the following corollary from the previous argument:
Let $H$ be a proper subgroup of $G$ then we may always find two distinct (linear) characters of $G$ that have the same restriction to $H$.
Indeed, by the previous argument there exists a conjugacy class $C$ of $G$ that does not intersect $H$. Let $D=GC$ and define $f$ to be the class function which is equal to $0$ on $D$ and $1$ on $C$ and let $g$ be the class function which is equal to $1$ everywhere.
Since $f$ and $g$ are (in a unique way) linear combinations of irreducible characters of $G$ and $fH=gH$ there must exist distinct irreducible characters $\psi_1$ and $\psi_2$ of $G$ which have the same restriction to $H$.

$\begingroup$ I do not think the last statement is correct. Take $G = A_{5}$ and $H$ to be a Sylow $5$subgroup of $G.$ No two distinct irreducible characters of $G$ agree on $H.$ The only two irreducible characters which have equal degree are the two irreducible characters of degree $3$, and these do not agree on a $5$cycle. $\endgroup$ Commented Oct 15, 2012 at 0:50

$\begingroup$ Hi Geoff, you are perfectly right, there is no reason to expect the two characters to be irreducible. $\endgroup$ Commented Oct 15, 2012 at 13:14

$\begingroup$ A followup question: math.stackexchange.com/q/3875776 . $\endgroup$– LSpiceCommented Oct 22, 2020 at 2:17
A superficially different counting argument, which boils down to the same proof as before:
If $H$ is a proper subgroup whose conjugates completely cover $G$, then let $G$ act on the right cosets of $H$ by right multiplication. This action is transitive. Since $H$ is a point stabilizer, the conjugates of $H$ are just all the point stabilizers. Then saying that the conjugates of $H$ cover $G$ is saying that every element of this permutation group has a fixed point. In a transitive permutation group, the average number of fixed points is $1$. The number of fixed points of the identity is the number of points, $[G:H]$. The only way every permutation can have at least the average number of fixed points is for every permutation to have exactly the average number of fixed points, so $[G:H]=1$ contradicting the assumption that $H$ is proper.

$\begingroup$ Actually, more is true. Let H be a proper subgroup of a finite group G. Not only is it true that some element of G lies in no conjugate of H, but in fact, there must be at least H such elements. This can be proved by a variation on the argument given in the comment by Harden. $\endgroup$ Commented Jan 31, 2012 at 19:15
A subgroup intersecting all conjugacy classes is usually called a conjugately dense subgroup. Here is a few related papers:
Levchuk, V. M. Sylow intersections and conjugately dense subgroups of Chevalley groups. (Russian) Algebra and linear optimization (Russian) (Ekaterinburg, 2002), 161–165, Ross. Akad. Nauk Ural. Otdel., Inst. Mat. Mekh., Ekaterinburg, 2002. 20G40
Zyubin, S. A.; Levchuk, V. M. Conjugately dense subgroups of locally finite Chevalley groups of Lie rank 1. (Russian) Sibirsk. Mat. Zh. 44 (2003), no. 4, 742748; translation in Siberian Math. J. 44 (2003), no. 4, 581–586.
Zyubin, S. A. Conjugately dense subgroups of 3dimensional linear groups over locally finite field. Internat. J. Algebra Comput. 15 (2005), no. 56, 1273–1280.
Zyubin, S. A. Conjugately dense subgroups of free products of groups with an amalgamated subgroup. (Russian) Algebra Logika 45 (2006), no. 5, 520537, 631; translation in Algebra Logic 45 (2006), no. 5, 296–305.
Zyubin, S. A. On irreducible conjugately dense subgroups of linear groups. (Russian) Dokl. Akad. Nauk 413 (2007), no. 4, 450453; translation in Dokl. Math. 75 (2007), no. 2, 266–269 20E06 (20G15)
Erfanian, Ahmad; Russo, Francesco Conjugately dense subgroups in generalized FCgroups. Acta Univ. Apulensis Math. Inform. No. 20 (2009), 79–91.