# Prove that the set of all functions is not a group under function composition.

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

You had the right idea. Consider this: Is there an inverse of the constant function $f$ defined by $f(1)=f(2)=f(3)=1$?
• @zzz, I think Hayden referred to the function $\;f(1)=f(2)=f(3)=1\;$. This thing doesn't have an inverse under composition. Nov 16, 2014 at 5:05
• @zzzz Yes, what Timbuc said is correct. It is the function defined by $f(1)=f(2)=f(3)=1$. Nov 16, 2014 at 14:03