Look here, the last post.
Hello there,
Can someone please explain to me the quickest method of finding out the Class equation of Dihedral groups, D_8, D_10, D_12 etc.
I am sure there is a quicker way than doing all the calculations, if someone can please tell me in ENGLISH what it is [since I can't understand the mathematical way]
I had initially thought there was a pattern, but it didnt work for D_8, since I thought if you have D_2n, then the class equation is
{1}, {x^k,x^m}, {x^w, x^v},....,{xy, ...., x^n-1 y}
where k+m = n, and w+v = n
This however was not the case for D_8, since {xy, ...., x^n-1 y} was split into 2, so if someone can please explain.
Thanks
I dont understand this part:
Can you please simplify the above if possible?The second step that I would do is choose an element not from the ones chosen, say . Now we will use the theorem which will make computations a little easier. The number of elements conjugate with has to be . Now thus it has at least four elements, by Lagrange's theorem it can therefore have elements it cannot be since is not in the center thus thus which means the number of elements conjugate with is just two. Now thus is conjugate to , we can stop here by the theorem thus is another conjugacy class.
The third step that I would do is choose an element not from the ones chosen, say . Now we will use the theorem which will make computations a little easier. The number of elements conjugate with has to be . Now thus it has at least three elements it cannot have all eight thus by Lagrange's theorem which implies that . Now thus is conjugate to , we can stop here by the theorem thus is another conjugacy class.
The centralizer of an element , denoted by , is the subgroup [tex]\{ a\in G|ax = xa\}. Think of it as a set of all elements which commute with , we denote it by .
The theorem (for finite groups) says that the number of elements in the same conjugacy class as is equal to .
Remember what means. It means the number of cosets of . It is in fact equal to .