How Many Elements Are in the Semigroup X^X?

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(Moderator's note: thread moved from "Linear & Abstract Algebra")

problem # 12).

Suppose X is a finite set with n elements. Show that the semigroup X^x has n^n elements.

I'm confused. Isn't semigroup a set of functions? So when it says n elements, it actually means n functions? Also what is X^x defined as?
 
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yaganon said:
problem # 12).

Suppose X is a finite set with n elements. Show that the semigroup X^x has n^n elements.

I'm confused. Isn't semigroup a set of functions? So when it says n elements, it actually means n functions? Also what is X^x defined as?

Well, if this semigroup is isomorphic to the set of functions from X to X, then n^n is clearly the right number. I looked at the semigroup page on Wikipedia and couldn't find that notation, though. Can't find it in your textbook?
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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