Math Help - Abstract problem - mappings and ops

1. Abstract problem - mappings and ops

Hi,

Could anyone help with the following question please?

Let S be a set with four elements, for example S={w,x,y,z}.
a) How many distinct operations can be defined on S?
b) How many commutative operations can be defined on S?
c) How many operations possessing an identity can be defined on S?
d) How many operations possessing an identity and such that every element has an inverse can be defined on S?

I'm thinking for part a) the answer is 4 to the power of 16 but should i be specifying the operation involved eg * or composition. The rest has me stumped. I usually just complete a Cayley table and work from there but if a) is 4 to the power of 16, I don't think I'll be doing that!

Thanks in advance for any help with this.

2. Originally Posted by jackiemoon
a) How many distinct operations can be defined on S?
b) How many commutative operations can be defined on S?
Let me do the first two to see if you can continue with the last two.

You got part (a) correct, think of a Cayley table, it consists of $4\cdot 4 = 16$ squares and for each square we can assign $4$ values from the set. Therefore, there are $4^{16}$ possible assignments. Exactly how you said!

To do part (b) again think of a Cayley table. Define the elements along the main diagnol i.e. define $w^2,x^2,y^2,z^2$. Once you do that define the elements in the upper half of this diagnol. But if you defined $wx$ (in the upper half) then $xw$ (in the lower half) is simply a reflection through the main diagnol. Therefore, the Cayley table is completely determined once you defined the main diagnol and the upper half. There are $10$ squares on the diagnol and upper half therefore $4^{10}$ possibilities for the number of commutative mappings.

This is Mine 11,4th Post!!!

3. Hey ThePerfectHacker,

Thanks for the help with this. I understand what you are saying about part b) but I'm not too sure about how to continue with parts c) and d) (well I think if I can understand c I should be able to get d). My thoughts on part c) are that if I choose say w as the identity element there is only one way I can compute the top row and column of the Cayley table ie.

w x y z
x
y
z

so they can only be completed in 1 possible way (?,) so that means there are 4 to the power of 9 ways to complete the table then. Am I on the right track with that?

Thanks again. I have an incredibly bad Abstract algebra lecturer and his assignments are my making my Sundays hell!

4. Originally Posted by jackiemoon

w x y z
x
y
z
!
There is a unique identity elements call it w. Then it means wx,xw,yw,wy, wz,zw are all determined.
The other ones are open to be determined.