> < ^ Date: Thu, 13 Nov 2003 22:51:04 -0500
> < ^ From: Bruce W. Colletti <bcolletti@compuserve.com >
> ^ Subject: Factors of a Permutation

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What is the command (if any) that returns (in a list) the disjoint =
cyclic factors of a permutation?  For instance, (1,2,3)(4,5,6) returns =
[(1,2,3), (4,5,6)]. =20

Building a routine is easy enough, but it's better to use a built-in =
one, if it's there.

Thanks.

Bruce
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<DIV><FONT face=3DArial size=3D2>What is the command (if any) that =
returns (in a=20
list) the disjoint cyclic factors of a permutation?&nbsp; For instance,=20
(1,2,3)(4,5,6) returns [(1,2,3), (4,5,6)].&nbsp; </FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Building a routine is easy enough, but =
it's better=20
to use a built-in one, if it's there.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Thanks.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Bruce</FONT></DIV></BODY></HTML>

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