# Talk:Conjugacy class

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## Centralizer is a subgroup

Reverting to previous text , as the centralizer is indeed a subgroup of G; if a,b in CG(g), then:

• CG is not empty, since (g*g)*g = g*(g*g).
• b * g = g * b → b -1 * g = g * b -1; thus b -1 is in CG(g)
• (a*b -1)*g = a*(b -1*g) = a*(g*b -1) = (a*g)*b -1 = (g*a)*b -1 = g*(a*b -1) → a*b -1 in CG(g).

thus by elementary group theory, CG(g) is a subgroup of G.

It is true that CG(g) is not a normal subgroup of G; but that's not what was stated in the article. Maybe this should be added to the article.User:chas_zzz_brown

## class equation

there should be written something about general (not only conjugacy) class equation; rather in group action than here, but here class equation redirects… konradek (talk) 15:29, 17 October 2008 (UTC)

## Example in conjugacy class equation

I think this example could be spiced a little bit... Let's say, proving that every group of order p2 is abelian is the next logical step, and quite trivial: the class equation for a hypothetical non-abelian group of order p2 is |G| = |Z(G)| + |H|, where H includes all elements of order p that are not in the center (of order p) - but this is absurd, because p (p - 1) does not divide p2 unless p = 2, and in this case there are only two groups of order 4 (both abelian). Albmont (talk) 16:49, 27 May 2009 (UTC)

(I don't know how to start a discussion/edit so here is my go:) I'm just a student but the sentence "Consider a finite p-group G (that is, a group with order p^n, where p is a prime number and n > 0). " is in contradiction with the definition of a p-group on wikipedia ( "p-group is a periodic group in which each element has a power of p as its order" )

so it's not |G| = p^n but for each g belongs to G, exists n belongs to N* such that |g| = p^n with p prime. — Preceding unsigned comment added by 67.212.8.144 (talk) 16:27, 6 December 2011 (UTC)

## Examples

The symmetric group S4, consisting of all 24 permutations of four elements, has five conjugacy classes. And you can compute that there are five conjugacy classes by taking the integer partition of 4. But how can you compute how many elements each conjugacy class has? PJ Geest (talk) 14:56, 1 June 2009 (UTC)

Sometimes the function of the partition is called "n!/z". If the partition has a1 1s, a2 2s, a3 3s, and a4 4s, then the conjugacy class has size 4!/( a1! * 1^a1 * a2! * 2^a2 * a3! * 3^a3 * a4! * 4^a4 ). For instance the partition 4 = 1 + 1 + 2 corresponds to the conjugacy class containing the permutation [1243] = (1)(2)(3,4) whose centralizer is Sym({1,2}) x < (3,4) > of order 2! * 2^1, so the conjugacy class has size 4!/4 = 6 = 4!/( 2! * 1^2 * 1! * 2^1 * 0! * 3^0 * 0! * 4^0 ). In a larger symmetric group, the partition 7 = 2+2+3 corresponds to the conjugacy class containing [2143675]=(1,2)(3,4)(5,6,7) whose centralizer is < (1,2),(3,4),(1,3)(2,4) > x <(5,6,7)> = <(1,2)> wr Sym( { {1,2}, {3,4} } ) x <(5,6,7)> of order 2^2*2! * 3^1, and so the conjugacy class has size 7!/( 0!* 1^0 * 2! * 2^2 * 1! * 3^1 * 0! * 4^0 * 0! * 5^0 * 0! * 6^0 * 0! * 7^0 ) = 7!/( 2^2*2! * 3^1 ) = 7!/24 = 210. JackSchmidt (talk) 19:53, 1 June 2009 (UTC)

## Error in class equation

Consider S_3 has 3 calsses C_1={e} , C_2={(acb),(bac),(cba)} , C_3 = {(cab),(bca)}. Center of S_3 is Z(S_3) = {e}. If we apply the equation exactly as it is written we'll get 6 = 1 + 6/1 + 6/3 + 6/2 and so 6=12 which is false.I think that sum should be a bit smaller. What do you think ? Stefan.petrea (talk) 17:00, 30 November 2009 (UTC)

If you exclude the conjugacy class of the identity element from that sum the identity becomes true(but that's not specified in the formula, is it so obvious ?) Stefan.petrea (talk) 17:06, 30 November 2009 (UTC)

I believe it is specified clearly and in a standard way. The sum is over the classes of size bigger than 1. The sum would be incorrect if it were over all non-identity classes; any abelian group gives a counterexample. JackSchmidt (talk) 17:41, 30 November 2009 (UTC)
Ok, then I suggest the formula be changed to ${\displaystyle |G|=|Z(G)|+\sum _{i=1\land [G:H_{i}]>1}^{i=|S|}[G:H_{i}]}$ , what do you think ? —Preceding unsigned comment added by Stefan.petrea (talkcontribs) 20:35, 30 November 2009 (UTC)