For any set we denote by the group of bijective maps from to

**Problem**. Let be any map which satisfies the following two conditions: and for all Let be a set with Prove that if is an abelian subgroup of then

**Solution**. The proof is by induction on If then Let be a set with and suppose that the claim is true for any set of size Let be an abelian subgroup of Clearly defines an action of on Let be the orbits corresponding to this action and consider two cases.

*Case 1*. : Fix an element Then Suppose that for some and let Then for some Thus, since is abelian, we have

Hence for all and thus So the stabilizer of is trivial and therefore, by the orbit-stabilizer theorem,

*Case 2*. : Let Clearly and, since we have for all For every and let the restriction of to and put

Then and is an abelian subgroup of Thus, by the induction hypothesis

for all Now, define by for all It is obvious that is one-to-one and so

**Remark**. The map defined by for all satisfies both conditions in the above Problem. So if and if is an abelian subgroup of then