# mar16 - and the result follows. Prove that | Aut ( Q 8 ) |...

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Math 4124 Wednesday, March 16 March 16, Ungraded Homework Exercise 4.4.1 on page 137 Prove that if σ Aut ( G ) and φ g is conjugation by g , then σφ g σ - 1 = φ σ ( g ) . Deduce that Inn ( G ) ± Aut ( G ) . Let x G . Then σφ g σ - 1 ( x ) = σ ( g ( σ - 1 ( x )) g - 1 ) = ( σ g )( σσ - 1 ( x ))( σ g ) - 1 = φ σ ( g ) ( x ) . Since this is true for all x G , it follows that σφ g σ - 1 = φ σ ( g ) . Furthermore Inn ( G ) Aut ( G ) (see page 134) and since every element of Inn ( G ) is of the form φ g for some g G , we conclude that Inn ( G ) ± Aut ( G ) . Exercise 4.4.4 on page 137 Prove | Aut ( Q 8 ) | ≤ 24. We write Q 8 in the usual way as 1 , ± i , ± j , ± k } . Since |h i i| = 4 and j / ∈ h i i , we see that Q 8 = h i , j i . Let θ be an automorphism of Q 8 . Since | i | = 4 and there are six elements of order 4, we see that there are at most six possibilities for θ i . Note that θ ( - i ) = - θ i , so θ j 6 = ± θ i . Since | j | = 4, we see that θ j also has order 4 and consequently there are at most four possibilities for θ j . Since θ is determined by its affect on { i , j } , we conclude there are at most 6*4 = 24 possibilities for
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Unformatted text preview: and the result follows. Prove that | Aut ( Q 8 ) | is a multiple of 4. We shall use the usual notation for the elements of Q 8 , that is { 1 , i , j , k } . Then Z ( Q 8 ) = { 1 } , so in particular | Z ( Q 8 ) | = 2. Therefore | Q 8 / Z ( Q 8 ) | = 4 and we deduce that | Inn ( Q 8 ) | = 4. Since Inn ( Q 8 ) Aut ( Q 8 ) , the result now follows from Lagranges theorem. Let G be a group of order 200. Prove that there exists H G such that 1 6 = H 6 = G . 200 = 2 3 * 5 2 . Therefore the number of Sylow 5-subgroups divides 8 and is congruent to 1 modulo 5. Therefore G has exactly one Sylow 5-subgroup, and so the Sylow 5-subgroup (which has order 25, so cannot be 1 or G ) is normal....
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