if n=45, calculate the number of cosets of An in Sn .
Your question makes no sense to me.
I'm not really sure about this, but I'll give it a go. By the way, what do you mean by cosets?
n=45
Solving for the number of (costs?) of An in Sn.
A45= S45
This is the most I can simplify it down to. If you post the full question from wherever you're getting it from, I may just be able to help.
Answer is 2..
From Lagrange's Theorem, we know that the number of cosets in An in Sn is [Sn:An] = |Sn|/|An| = 2
Note An = n!/2 = |Sn|/2
To calculate the number of cosets of An (alternating group of even permutations) in Sn (symmetric group of all permutations), we need to find the index of An in Sn.
The index of a subgroup H in a group G is defined as the number of distinct left (or right) cosets of H in G. We can calculate the index using the formula:
Index (G : H) = |G| / |H|
Here, |G| represents the order (or size) of G, and |H| represents the order of H.
In this case, we want to calculate the number of cosets of An in Sn. Since An is a proper subgroup of Sn, we know that the size of An is less than the size of Sn.
The order of Sn, denoted as |Sn|, is equal to n! (n factorial), which represents the number of permutations of n elements.
Therefore, |Sn| = n!
For the order of An, denoted as |An|, we use the formula:
|An| = n!/2
This is because the alternating group An contains only half the number of permutations of Sn, since it consists of even permutations (those that can be written as an even number of transpositions).
Therefore, |An| = n!/2.
Now, based on the formula for calculating the index, we can substitute the values:
Index (Sn : An) = |Sn| / |An|
= n! / (n!/2)
= n! * 2 / n!
= 2
So, there are 2 cosets of An in Sn, which means that the index of An in Sn is 2.