Generating a finite group from elements in each conjugacy class - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:47:18Z http://mathoverflow.net/feeds/question/26979 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class Generating a finite group from elements in each conjugacy class Jamie Vicary 2010-06-03T21:59:31Z 2012-10-15T13:16:26Z <p>Is there a finite group such that, if you pick one element from each conjugacy class, these don't necessarily generate the entire group?</p> http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class/26982#26982 Answer by Richard Stanley for Generating a finite group from elements in each conjugacy class Richard Stanley 2010-06-03T22:14:01Z 2010-06-04T14:33:59Z <p>Very dumb mistake. Answer withdrawn.</p> http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class/26992#26992 Answer by Steve D for Generating a finite group from elements in each conjugacy class Steve D 2010-06-04T00:09:02Z 2010-06-04T00:09:02Z <p>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.</p> <p>Steve</p> http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class/26993#26993 Answer by David Speyer for Generating a finite group from elements in each conjugacy class David Speyer 2010-06-04T00:21:36Z 2010-06-04T00:31:01Z <p>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 <code>$g_1 H g_1^{-1} = g_2 H g_2^{-1}$</code>, 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.</p> <p><strong>UPDATE:</strong> I found a reference. According to <a href="http://www.ams.org/journals/bull/2003-40-04/S0273-0979-03-00992-3/home.html" rel="nofollow">Serre</a>, this result goes back to Jordan, in the 1870's.</p> http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class/82256#82256 Answer by DavidLHarden for Generating a finite group from elements in each conjugacy class DavidLHarden 2011-11-30T08:04:11Z 2011-11-30T08:04:11Z <p>A superficially different counting argument, which boils down to the same proof as before: </p> <p>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.</p> http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class/95817#95817 Answer by Hugo Chapdelaine for Generating a finite group from elements in each conjugacy class Hugo Chapdelaine 2012-05-02T23:22:33Z 2012-10-15T13:16:26Z <p>The impossibility also follows from <strong>Jordan's lemma</strong>:</p> <p><em>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).</em></p> <p>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.</p> <p>Note also that one gets the following corollary from the previous argument: </p> <p><em>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 $H$.</em></p> <p>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 <strike>irreducible </strike> characters $\chi$ and $\psi$ of $G$ which have the same restriction to $H$.</p>