## Generating a finite group from elements in each conjugacy class

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

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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. – Jamie Vicary Jun 3 2010 at 22:05
I don't see why there should be a connection: it's certainly not the case for finite cyclic and finite dihedral subgroups of $SO(3)$. – Victor Protsak Jun 4 2010 at 0:05

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.

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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 well-known to never happen.

Steve

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"Well-known" as in "an easy exercise". – Steve D Jun 4 2010 at 0:12
Well-known to those who know it, I suppose. What (precisely) are saying never happens? – Kevin O'Bryant Jun 4 2010 at 0:14
It's worth pointing out that this argument really uses finite. For example, for compact simple Lie groups every element lies in some torus. – Noah Snyder Jun 4 2010 at 0:20
Very minor nitpick, you mean $(|M|-1)[G:M]+1 < |G|$ – Noah Snyder Jun 4 2010 at 0:22
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. – Geoff Robinson Apr 28 2011 at 19:32
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The impossibility also follows from Jordan's lemma:

Let $G$ act transitively on a set $\Omega$ with $|\Omega|:=n\geq 2$ then there exists a $g\in G$ such that $\chi(g)=0$ (here $\chi(g)$ denotes the permutation character).

Here $\chi(g)$ denotes the permutation character. 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 is negative. So let $H$ be the group generated by the representatives of each conjugacy class of $G$ and assume 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 there exists a $x\in G$ such that $x g_i H\neq g_i H$ for each left coset $g_i H$. 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 $g^{-1}xg\notin H$. 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 on $H$.

Indeed, by the previous argument there exists a conjugacy class $C$ of $G$ that does not intersect $H$. Let $D=G-C$ 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 $f|H=g|H$ there must exist distinct irreducible characters $\chi$ and $\psi$ of $G$ which have the same restriction to $H$.

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 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. – Geoff Robinson Oct 15 at 0:50 Hi Geoff, you are perfectly right, there is no reason to expect the two characters to be irreducible. – Hugo Chapdelaine Oct 15 at 13:14

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There are 5 conjugacy classes in the quaternion group (8=1^2+1^2+1^2+1^2+2^2). Also -jij=-jk=-i so i and -i are in the same conjugacy class – Noah Snyder Jun 3 2010 at 22:38
The question is equivalent to having a maximal subgroup which intersects every conjugacy class. This is clearly impossible in a p-group, since maximal subgroups are normal. – Steve D Jun 3 2010 at 23:41
A meta-remark: it seems to me wrong to vote down incorrect answers just because they are incorrect. Of course, an answer can be plain silly, but this is not such an answer. I don't obsessively check my answers on Mathoverflow because I know that if I do get something wrong it will be picked up. – gowers Jun 4 2010 at 14:33
Someone also voted my answer down, and it is correct! – Steve D Jun 4 2010 at 14:54
I think it can be good to vote down wrong answers so that they're below the correct answers. However, once they're below the correct answer there's no need to continue to pile on. (I didn't vote this question down, and instead voted both other questions up.) – Noah Snyder Jun 4 2010 at 15:38

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.

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 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. – Marty Isaacs Jan 31 2012 at 19:15