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July 15th, 2012, 12:26 PM   #1
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conjugacy classes, center, alternating group etc....

Ok can anyone give me one good example that contains all this? i know the definitions but i cant apply it because i have no examples to work with. Specifically for , can anyone show me how to find the center, orbit, and the conjugacy classes of this group? also how about the ?

i really need someone to show me a clear step on how to finding these because i cant proceed without knowing how to find the centers, orbit etc... i would really appreciate the help
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July 16th, 2012, 02:06 AM   #2
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Re: conjugacy classes, center, alternating group etc....

The symmetric group is non abelian; so, , the center, is trivial since

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July 17th, 2012, 12:44 AM   #3
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Re: conjugacy classes, center, alternating group etc....

it is NOT true that a group that is non-abelian has a trivial center.

Consider, for example, the group D4 =

{e, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 3), (2 4), (1 2)(3 4), (1 4)(2 3)}

this group has center Z(D4) = {e, (1 3)(2 4)}.

********
for Sn, the conjugacy classes are just the cycle types. for S4 these are:

{e} = Z(S4) (one cycles)
{(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)} (transpositions)
{(1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 3 4), (2 4 3)} (3-cycles)
{(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} (double 2-cycles)
{(1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)} (4-cycles)

the fact that all cycle types are conjugate to any other of the same type stems from the fact that:

if ? = (a1 a2....ak) (a k-cycle), then for any permutation ?:

???^-1 = (?(a1) ?(a2).....?(ak)) (replace the cycle elements of ? with their images under ?).

for An, the situation is more interesting. although any two elements conjugate in An are still conjugate in Sn, the reverse is not necessarily true. the reason being, we have "fewer things to conjugate with", that is, the "?" have to live in An as well.

so which 3-cycles are conjugate to each other? for example, are (1 2 3) and (1 2 4) conjugate? note that for (1 2 3) and (1 2 4) to be conjugate, we need a permutation ? with:

?(1) = 1
?(2) = 2
?(3) = 4

but this forces ?(4) = 3 (since ? must be bijective), that is: ? = (3 4), which isn't IN A4 (actually we should check that (1 2 3) is not conjugate to (2 4 1) or (4 1 2), as well, which i encourage you to do).

how about (1 2 3) and (1 3 2)? we would need:

?(1) = 1
?(2) = 3
?(3) = 2
?(4) = 4, which is again a transposition. maybe (1 2 4) and (1 3 2) are conjugate? we start with:

?(1) = 1
?(2) = 3
?(4) = 2

which means ?(3) must be 4, so ? = (2 3 4), which IS in A4. that is:

(2 3 4)(1 2 4)(2 3 4)^-1 = (1 3 2).

convince yourself that the 3-cycles split into TWO conjugacy classes in A4:

{(1 2 3), (1 3 4), (1 4 2), (2 4 3)} and
{(1 2 4), (1 3 2), (1 4 3), (2 3 4)}

finally, note that:

(1 2 3)(1 2)(3 4)(1 2 3)^-1 = (2 3)(1 4) = (1 4)(2 3) (because disjoint cycles commute), and
(1 2 4)(1 2)(3 4)(1 2 4)^-1 = (2 4)(3 1) = (1 3)(2 4), so the double 2-cycles remain all conjugate in A4.
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