1. The set $S(X,X)$ of all mappings of a set $X$ to itself with the composition of mappings in the role of multiplication, where $|X|>1$. Why is not it a group?
  2. Let $X$ be a nonempty set. Then the idempotents of the semigroup $S(X,X)$ of all mappings of $X$ to itself are precisely the mappings $f: X \to X$ satisfying the condition $f(x)=x$ for every $x \in f(X)$. An element $x$ of a semigoup is callled an idemotent if $xx=x$. My question is this: Here why $f(x)=x$ for any $x\in f(X)$?

Thanks a Lot!

  • 1
    $\begingroup$ I will be very frank: given the mathematical level of the questions you have asked here on the site before, I would have assumed you could easily answer these questions yourself. $\endgroup$ – Zev Chonoles Sep 22 '13 at 6:45
  • 1
    $\begingroup$ @ZevChonoles: I'm a beginner of abstract-algebra and I want to be sure of something which I'm not sure. This is the purpose that I post question which seems that a little stupid. $\endgroup$ – Paul Sep 22 '13 at 6:53


  1. If $|X|>1$, then there exist functions $f:X\to X$ without inverses.

  2. Write out what it would mean to have $f\circ f=f$.

  • $\begingroup$ 1, Why $f$ has not inverse? 2, I only have $f(f(x))=f(x)$. $\endgroup$ – Paul Sep 22 '13 at 6:30
  • $\begingroup$ Regarding 2: Saying that $$f(f(x))=f(x)$$ for all $x\in X$ is the same thing as saying that $$f(y)=y$$ for all $y\in f(X)$. $\endgroup$ – Zev Chonoles Sep 22 '13 at 6:32
  • $\begingroup$ Regarding 1: There will be some functions $f:X\to X$ that do not have inverses, and others that do. Do you understand what it means for a function to be injective? Do you see how can make a function $f:X\to X$ that is not injective if $X$ has more than one element? $\endgroup$ – Zev Chonoles Sep 22 '13 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.