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Simple groups can be characterised by following property:

A group $G$ is simple if and only if for any $1\neq x\in G$, the conjugacy class of $x$ in $G$ generates the whole group $G$.

The question I am interested is natural one related to above fact?

Q. If $G$ is a group such that for any $x\in G$, the conjugacy class of $x$ in $G$ generates a proper subgroup, what can we say about $G$? (nilpotent, solvable, supersolvable, or else?) Is there characterization (or classification) of such groups?

Example: Let $G$ be any non-abelian finite $p$-group. Then for $x\in G$, there is a maximal subgroup $M$ of $G$ containing $x$; then the subgroup generated by the conjugacy class of $x$ is contained in $M$, hence it is proper.

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$A_5 \times C_2 \times C_2$ has this property, so you cannot deduce solvability. –  Derek Holt Jul 5 '13 at 6:32

1 Answer 1

Not quite done, but time has been scarce lately:

Proposition: In a finite group with non-cyclic abelianization, the subgroups generated by each conjugacy class are proper.

Proof: Suppose $G$ is a finite group whose abelianization $G/[G,G]$ is not cyclic. Let $x \in G$. Then $G/[G,G] \neq \langle x \rangle [G,G]/[G,G]$, so there is some maximal subgroup $M/[G,G]$ such that $\langle x \rangle [G,G]/[G,G] \leq M/[G,G]$. Since $G/[G,G]$ is abelian, $M/[G,G]$ is normal, and so $M \lhd G$ is normal, and so contains the subgroup generated by all conjugates of $x$. $\square$

Proposition: If $G/[G,G]$ is cyclic and $G$ is finite (or has a composition series), then there is a conjugacy class that generates $G$ as a subgroup.

Proof: This is a special case of theorem 4.7 of Berrick–Robinson (1993). $G$ itself has finite composition length, so Robinson's result says that $G$ is the subgroup generated by a single conjugacy class if and only if $G/[G,G]$ is. However, $G/[G,G]$ is generated by a single element (=conjugacy class for abelian groups) if and only if it is cyclic. $\square$

References:


Proof: (Of conjecture, my attempt) Suppose $G/[G,G]$ is cyclic, generated by $x[G,G]$, and set $N$ to be the subgroup generated by the conjugates of $x$. Let $g \in G$. Then $x^g = x[x,g]$ so $[x,g] \in N$. Hence $x \in Z(G/N)$. ...

Counterexample: If $G/[G,G] = \langle x \rangle [G,G]/[G,G]$, then $G$ need not be the subgroup generated by the conjugacy class of $x$.

Proof: Take $G = \langle x \rangle \times Y$ with $Y$ non-identity perfect. Then the conjugacy class of $x$ is just $x$ itself, a proper subgroup. $\square$

Fix: the only problem is when $x$ is contained in a maximal normal subgroup not containing $[G,G]$, but in that case the quotient is non-abelian simple, and so a joe random element $y$ of that quotient should be such that $G$ is the subgroup generated by the conjugacy class of $xy$.

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Oh and one more todo: for infinite groups, check if quasi-cyclic abelinizations are the only obstruction. –  Jack Schmidt Jul 5 '13 at 16:43
    
interesting! Proposition can be applied to finite $p$-groups, and also to the example by Derek Holt. But, sorry, I do not understand the conjecture (please, just restate it). –  Marshal Kurosh Jul 6 '13 at 3:56
    
Fixed. If G is perfect and contained in GAP's library, then the conjecture is true. I haven't had a chance to fix the conjecture's “proof” to actually be a proof. –  Jack Schmidt Jul 6 '13 at 5:59
    
Where this conjecture appeared first? (or who made this conjecture?) –  Marshal Kurosh Jul 6 '13 at 6:05
2  
I made this conjecture. –  Jack Schmidt Jul 6 '13 at 12:43

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