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This conjecture in "Unsolved problems in group theory" No.18: 9.24:

Conjecture: every finite simple non-abelian group $G$ can be represented in the form $G=CC$, where $C$ is some conjugacy class of $G$.

I want to prove this conjecture is right for simple group $A_n$ in $S_n$ ($n>4$), but I have no idea how to deal with it. I hope someone could give me some advice, or some old results about this.

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    $\begingroup$ What do you mean $A_{n}$ in $S_{n}$? The simple group is $A_{n}$. The problem is to prove that if $G$ is a finite simple group, there is a conjugacy class $C$ such that every element of $G$ can be expressed in the form $ab$ with $a,b \in C.$ The class $C$ has to be closed under taking inverses. The conjecture has been checked for many simple groups, The requirement is stronger than asking that every element of $G$ is a commutator. $\endgroup$
    – Geoff Robinson
    Commented Oct 17, 2016 at 1:08
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    $\begingroup$ For $n\ge5$ odd, is every element of $A_n$ a product of two $n$-cycles? $\endgroup$
    – Gerry Myerson
    Commented Oct 17, 2016 at 5:51
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    $\begingroup$ @GerryMyerson For $n=5$, the answer to your question is yes. On the other hand, note that $n$-cycles fall in two distinct conjugacy classes in $A_n$ when $n$ is odd. For $n=5$, we don't have $A_5=CC$ where $C$ is one of these classes. $\endgroup$
    – verret
    Commented Oct 17, 2016 at 6:29
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    $\begingroup$ Before you can prove it, you will need to identify the class $C$ for which this might work. What I would do would be to start by doing some computer experiments for $n=5,6,7,8,9,\ldots$ in order to find out for which $C$ the conjecture is true and, based on that, try to formulate a conjecture, $\endgroup$
    – Derek Holt
    Commented Oct 17, 2016 at 8:49
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    $\begingroup$ As I mentioned, the class $C$ must be closed under taking inverses. It follows easily from this that any product of the form $ab$ with $a,b \in C$ is a commutator ( for $a = c^{-1}b^{-1}c$ for some $c \in G$). Hence Thompson's conjecture implies that every element of $G$ is a commutator when $G$ is simple, which is a (now proved) conjecture of Ore. $\endgroup$
    – Geoff Robinson
    Commented Oct 17, 2016 at 9:36

1 Answer 1

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This was proved in this paper:

MR0183763 (32 #1241) Reviewed 
Xu Cheng-hao
The commutators of the alternating group. 
Sci. Sinica 14 1965 339–342. 
20.20

My institution does not have an electronic copy, so there you are on your own.

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    $\begingroup$ See also Edward Bertram, Even permutations as a product of two conjugate cycles, J Comb Thy Ser A 12 (May 1972) 368-380, sciencedirect.com/science/article/pii/0097316572901021 $\endgroup$
    – Gerry Myerson
    Commented Oct 17, 2016 at 21:56
  • $\begingroup$ That is great. It is useful for me. Thank you! $\endgroup$
    – C. Simon
    Commented Oct 18, 2016 at 11:35

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