Dummit and Foote Section 4.4 Automorphisms, Exercise 3 reads as follows:
Prove that under any automomorpism of , r has at most two possible images and s has at most 4 possible images. Deduce that |Aut()| 8
In a post on Project Crazy Project nbloomf gives a proof:
First the following Lemma is stated and proved:
Lemma: Let be a group homomorphism. Then , with equality if is bijective
Then the proof for the D&F exercise follows:
Let be an automorphism of . Automorphisms must preserve order, so that has order 4 and has order 2. There are 2 elements of order 4 in , so that we have at most 2 choices for . Similarly, there are 5 elements of order 2 in . However, one of these is in . By the lemma, we have at most 4 choices for
I cannot follow the reassoning:
Similarly, there are 5 elements of order 2 in . However, one of these is in . By the lemma, we have at most 4 choices for
Can anyone please give a much more explicit proof of this assertion - why does one of the elements being in mean that we have at most 4 choices for ???
Appreciate some help.