# ahw7 - Math 4124 Wednesday March 30 Seventh Homework...

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Math 4124 Wednesday, March 30 Seventh Homework Solutions 1. Exercise 4.4.3. Prove that under any automorphism of D 8 , r has at most two possible images and s has at most four possible images. Deduce that | Aut ( D 8 ) | ≤ 8. Here D 8 = { r , s | r 4 = s 2 = 1 , rs = sr - 1 } . As usual, D 8 = { 1 , r , r 2 , r 3 , s , sr , sr 2 , sr 3 } . Here | r | = | r 3 | = 4 and | r 2 | = | s | = | sr | = | sr 2 | = | sr 3 | = 2. Let α Aut D 8 . Since | r | = 4, we see that | α r | = 4 (the order of an element is preserved by an isomorphism, so in particular it is preserved by an automorphism), hence α r = r or r 3 ; in other words there are at most two possible choices for α r . Next | s | = 2, hence | α s | = 2, so it would appear at ﬁrst sight that there are 5 possible choices for α s , namely r 2 , s , sr , sr 2 , sr 3 . However, we can rule out the case α s = r 2 . Indeed we have α r = r or r 3 (from the ﬁrst paragraph), hence α r 2 = ( α r ) 2 = r 2 or r 6 and we see that

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ahw7 - Math 4124 Wednesday March 30 Seventh Homework...

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