This exercise shows that for every automorphism of is inner. Fix an integer with .

- Prove that every automorphism of a group permutes the conjugacy classes of . That is, if is a conjugacy class and an automorphism of , then is a conjugacy class of .
- Let be the conjugacy class of transpositions in and let be the conjugacy class containing any element of order 2 in that is not a transposition. Prove that . Deduce that any automorphism of sends transpositions to transpositions. [Hint: Use §4.3 #33.]
- Prove that for each , , , …, for some distinct integers .
- Show that , , …, generate and deduce that any automorphism of is uniquely determined by its action on these elements. Use (c) to show that has at most automorphisms. Conclude that for .

- [Short way]
Let be an automorphism of , and let be a conjugacy class. Note that , so that . In particular, is a conjugacy class of .

[Long way]

Let and let be a conjugacy class. We claim that is a conjugacy class. Let . Then and for some . Since is a conjugacy class, there exists an element such that ; then , so we have . Thus and are conjugate.

Let and and suppose that and are conjugate; then for some , . Now since is surjective, we have , , and for some , , and . Now , and since is injective, we have . Thus and are conjugate, a contradiction since is a conjugacy class.

Thus is a conjugacy class for all automorphisms and conjugacy classes . Moreover, we have and , so that acts on the set of conjugacy classes of by function application.

- We now prove some lemmas.
Lemma 1: If , then . Proof: We proceed by induction on . For the base case () note that . For the inductive step, suppose that the conclusion holds for some . Now note that the polynomial has roots and opens upward; thus for , . Now , so that . Thus . By induction, the conclusion holds for all .

Lemma 2: Suppose . If , then . Proof: Note that , so that . Hence , so that . Thus .

Lemma 3: If and , then . Proof: We proceed by induction on . For the base case (), there are three possibilities for . If , we have . If , we have . For the inequality holds by Lemma 1. Now for the inductive step, suppose that for some the inequality holds for all . If is even, then if and only if . Thus by Lemma 2, for all . If is odd, then if and only if or . In the first case, use Lemma 2 as before. In the second case, the inequality holds by Lemma 1. Thus for all , we have . By induction, the inequality holds for all and all .

Lemma 4: If and , then . Proof: If , then . Now . If , then . Now . If , then or . If , then . If , then . .

Now we move to the main result. Note that if is 2 or 3, then all elements of order 2 are transpositions; thus any automorphism of sends transpositions to transpositions. Now if and is an element of order 2, then is a product of disjoint 2-cycles. If is not a transposition, then ; note then that . By §4.3 #33, the number of transpositions in is , and the number of conjugates of is . These two numbers are not equal for and all , since otherwise we have , which contradicts Lemmas 3 and 4. Now if is an automorphism of and the conjugacy class of transpositions, then is a conjugacy class of elements of order 2. If , we have a contradiction since . Thus every automorphism of maps transpositions to transpositions if .

- We begin with some lemmas.
Lemma 5: Let . Then for all , if and only if . Proof: if and only if since is injective, if and only if since is a homomorphism.

Lemma 6: Let with . Then . Proof: For each , is a transposition by part (b). Thus . Suppose ; then for some , we have . But then and are disjoint and thus commute, a contradiction since and do not commute. Now suppose ; since each is a transposition, we have for all and , a contradiction since is injective and (since ) there exist with . Thus .

Now to the main result. If , then has only the trivial automorphism, under which . If , then by part (b) each of is a transposition for , and by Lemma 6 the move a common element . Since is a bijection, the second element moved by the are pairwise distinct. Thus for some distinct we have , , …, .

- In §3.5 #3, we saw that is generated by . Note that . Thus , and so any automorphism of is uniquely determined by its action on these transpositions. By part (c), there are at most possible choices for the images of the under an automorphism (one for each assignment of the to the integers . Thus for .

## Comments

How about this for part 1: Let be a conjugacy class in . Then for each , .

I like it!

Some typos: I don’t know what is “ell” in Lemma6, and “with \latex k\neq l$” also looks like a miss.

Thanks!