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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 '10 at 22:05
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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 '10 at 0:05

4 Answers 4

up vote 46 down vote accepted

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 '10 at 0:12
    
Well-known to those who know it, I suppose. What (precisely) are saying never happens? –  Kevin O'Bryant Jun 4 '10 at 0:14
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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 '10 at 0:20
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Very minor nitpick, you mean $(|M|-1)[G:M]+1 < |G|$ –  Noah Snyder Jun 4 '10 at 0:22
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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 '11 at 19:32

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 '12 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 '12 at 13:14

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 '12 at 19:15

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