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image of ’is the (additive) cyclic subgroup hai Z=nZ. Then since ’is surjective we must have hai= Z=nZ. By the Lemma, this happens if and only if aand nare coprime, i.e. a2(Z=nZ) , in which case the map ’(x) = axis also invertible with inverse ’ 1(x) = a 1x.

Z=nZ !Aut(Z=mZ) ˘=(Z=mZ) and the order nelement 1 in the LHS will have order dividing both nand the cardinality ’(m) of the automorphism group. Thus is has order 1 and so ˚is the trivial homomorphism. 3. S n ˘=A noZ=2Z. 4. The identity morphism (Z=nZ) !Aut(Z

15/5/1992· S_n -> Aut (S_n) and this is 1-1 except for n=2, because in every other case the. center of S_n is trivial. (By the way this means that S_2 is not equal. to Aut (S_2).) So proving "Clegg''s conjecture" amounts to showing that. this homomorphism is onto, which apparently is true unless n is 6.

15/5/1992· S_n -> Aut (S_n) and this is 1-1 except for n=2, because in every other case the. center of S_n is trivial. (By the way this means that S_2 is not equal. to Aut (S_2).) So proving "Clegg''s conjecture" amounts to showing that. this homomorphism is onto, which apparently is true unless n is 6.

MATH 3005 Homework Solution Han-Bom Moon Homework 6 Solution Chapter 6. 1.Find an isomorphism from the group of integers under addition to the group of even integers under addition. Let 2Z be the set of all even integers. Deﬁne a map ˚: Z !2Z as ˚(n) = 2n.

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