**Notation 1**. For a group we will denote by and the center of , the group of automorphisms of and the group of inner automorphisms of respectively.

**Notation 2**. By we will mean the map defined by for all Recall that, by definition,

**Remark**. For every and we have

**Definition**. A group is called **complete** if and i.e. every automorphism of is inner.

**Problem 1**. Let be a group with Prove that the centralizer of in is trivial. In particular,

**Solution**. Let and Let be in the centralizer of in So for every we have Thus for all Hence for all Therefore and so

**Problem 2**. Let be a group with Prove that if is a characteristic subgroup of then is complete.

**Solution**. Let and So, by Problem 1 and the fact that is a subgroup of we only need to prove that Let Since is characteristic in for every there exists some such that Define by It is easy to see that So

for all Let

and let Then by the above remark and

Thus, by the above remark and

We can re-write as which means that is in the centralizer of Thus, by Problem 1, and so for all Thus and hence

**Problem 3**. Prove that if is a non-abelian simple group, then is complete.

*Proof*. Let and Since is a normal subgroup of and is non-abelian and simple, By Problem 2, we only need to prove that is characteristic in So let We need to prove that Let

Clearly is normal in because is normal in We also have

which implies that is simple, because is simple. So either or

*Case 1*. Then and we are done.

*Case 2*. Then will be in the centralizer of because both and are normal in Hence, by Problem 1, and so which is absurd because is non-abelian.