Consider the set $F$ of all functions from $\{1,2,3\}$ to $\{1,2,3\}$. There are $3^3= 27$ of them.
Prove this set is not a group under function composition.

I thought that it violates the inverse element property, but not sure how. I believe identity in our case is the identity map. Not really sure how to show an example how it fails under inverse. Help much appreciated


1 Answer 1


You had the right idea. Consider this: Is there an inverse of the constant function $f$ defined by $f(1)=f(2)=f(3)=1$?

  • $\begingroup$ are you referring to 1 in another set? this function does not exist $\endgroup$ Nov 16, 2014 at 3:22
  • $\begingroup$ @zzz, I think Hayden referred to the function $\;f(1)=f(2)=f(3)=1\;$. This thing doesn't have an inverse under composition. $\endgroup$
    – Timbuc
    Nov 16, 2014 at 5:05
  • $\begingroup$ @zzzz Yes, what Timbuc said is correct. It is the function defined by $f(1)=f(2)=f(3)=1$. $\endgroup$
    – Hayden
    Nov 16, 2014 at 14:03

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