hw1newsoln

# hw1newsoln - MATH 150B WINTER 2010 HOMEWORK 1 SOLUTIONS...

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MATH 150B WINTER 2010 HOMEWORK 1 SOLUTIONS ADAM SORKIN Many thanks to Jeﬀ Ferreira for assistance on this homework, and, I’m sure, many more. 5.8.2 Let S be a group on which G operates, and let H = { g G | gs = s for all s S } . Prove that H is a normal subgroup of G . Proof. This is a quick computation. Let h H,g G,s S . Then ghg 1 ( s ) = g ( h ( g 1 ( s )) = g ( g 1 ( s )) by deﬁnition of h = s. Hence if h H , then for all g G we have ghg 1 H , and so H is normal. ± 5.8.3 Let G be the dihedral group of symmetries of a square. Is the action of G on the vertices a faithful action? on the diagonals? Proof. For the action of G on the vertices, we do have a faithful action. This is because only the identity element in D 4 ﬁxes all the vertices simultaneously. For the action of G on the diagonals, we do not have a faithful action. This is because, for example, rotation by π ﬁxes both diagonals. ± 5.8.5 A group G operates faithfully on a set S of ﬁve elements, and there are two orbits, one of order 3 and one of order 2. What are the possibilities for G ? Proof. Recall that a G -action on S is equivalent to a homomorphism ρ : G Perm( S ). Because the action splits S into 2 orbits, O 3 and O 2 of size 3 and 2 respectively, the homomorphism splits into a product ρ = ( ρ 1 2 ): G Perm( O 3 ) × Perm( O 2 ) (cf 5.8.4). That is, ρ is a morphism to S 3 × S 2 . Furthermore, this mapping is injective, as the action is faithful. Hence G is (isomorphic to) a subgroup of S 3 × S 2 , subject to the condition that G has some elements of orders 2 and 3. One quickly sees there are only two subgroups of S 3 × S 2 satisfying this condition, namely S 3 × S 2 itself, and the subgroup ((123) , (12)) isomorphic to the cyclic group of order six. ± 5.Misc.2bc (b) Prove that the map φ : G Aut( G ) deﬁned by g 7→ (conjugation by g ) is a homomorphism, and determine its kernel. Proof. Veriﬁcation that φ is a morphism requires checking the maps pointwise. Letting φ ( g ) = φ g : G G , one has that φ gh ( k ) = gh ( k )( gh ) 1 = g ( hkh 1 ) g 1 = g ( φ h ( k )) g 1 = φ g ( φ h ( k )). Hence φ gh = φ g φ h . The kernel of φ is { g G | ghg 1 = h for all h G } , which is precisely Z ( G ), the center of G . ± (c) The automorphisms which are conjugation by a group element are called inner automor- phisms . Prove that the set of inner automorphisms, the image of φ , is a normal subgroup of Aut( G ). 1

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2 ADAM SORKIN Proof. Consider some inner automorphism φ g and an arbitrary automorphism ψ . By direct computation pointwise, one has ψ φ g ψ 1 = φ ψ ( g ) , whence φ ( G ) is a normal subgroup of Aut( G ). ± 5.Misc.7 Let G be a ﬁnite group acting on a ﬁnite set S . For each element g G , let S
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## This note was uploaded on 05/19/2010 for the course MAT150B math150b taught by Professor Vazirani during the Winter '10 term at UC Davis.

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hw1newsoln - MATH 150B WINTER 2010 HOMEWORK 1 SOLUTIONS...

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