00:01
Hi here for the given question.
00:03
We need to prove that for n greater than or equal to 4 then the center of alternating subgroup.
00:18
An is a tribal subgroup.
00:27
What is the value of then we need to find what is the value of z of a n for n equals to 0 comma 1 comma 2 comma 3.
00:44
So here now in order to prove this let us consider an orbital element sigma in a n now here we need to show that for every for every tau in a n sigma tau is equal to tau sigma.
01:09
So here we know that sigma and tau are disjoint cycles.
01:14
They are disjoint cycles meaning they don't share any element in a common.
01:22
So here further let us consider a case where sigma and tau have a common element without loss of generality.
01:31
We will assume that the common element is one let one be a common element.
01:38
So here we can write sigma is equal to 1 comma a 1 comma a 2 up to a k and tau can be written as 1 comma b 1 comma b 2 up to bm.
01:55
So here now we need to prove that the value of sigma tau is equal to tau sigma.
02:02
So here we shall consider the two cases below the case one says that k not equal to m if the value of k is not equal to m then here tau and sigma are the disjoint cycles of the different length are disjoint cycle of different length.
02:29
So here we can say that for this case, we will always get sigma tau is equal to tau sigma.
02:36
Now further let us consider an another case where the value of k is equal to m.
02:42
So here the length of the cycle is same which means length of two cycle is same.
02:51
Now here we need to show this value of sigma tau is equal to tau sigma by considering the action of sigma and tau on each and every element.
03:05
So here let us consider an arbitrary element in x other than one since we know that this both have the same length.
03:12
So the number of fermentation on both may or may not be same but they have a different orders...