Dummit and Foote Section 4.4 Automorphisms, Exercise 3 reads as follows:

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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

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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

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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.

Peter