New PDF release: Abstract Algebra I

By Randall R. Holmes

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Finally, a0 = e ∈ H. Therefore, am ∈ H for all m ∈ Z, whence a ⊆ H. 4 Order of element Let G be a group and let x ∈ G. If xn = e for some positive integer n, then the least such integer is the order of x, written ord(x). If no such positive integer exists, then x has infinite order, written ord(x) = ∞. 1 In the multiplicative group C× , find Example (a) ord(i) (i = √ −1), (b) ord(3). Solution In this group, 1 plays the role of e. (a) Since i1 = i i2 = −1 i3 = i2 i = −i i4 = i3 i = −i2 = 1 = e, it follows that ord(i) = 4.

Let x , y ∈ G . ) Since ϕ is surjective, we have x = ϕ(x) and y = ϕ(y) for some x, y ∈ G. Then x y = ϕ(x)ϕ(y) = ϕ(xy) (homomorphism property) = ϕ(yx) (G is abelian) = ϕ(y)ϕ(x) (homomorphism property) =yx. Therefore, G is abelian. (⇐) Assume that G is abelian. Since G ∼ = G we have G ∼ = G as well. Therefore the proof above shows that G is abelian. Recall that the symmetric group S3 has order 3! = 6. Since Z6 also has order 6, there exists a bijection from S3 to Z6 , so one might wonder whether these two groups are actually isomorphic.

N − 1)1} = {0, 1, 2, 3, . . , n − 1} = Zn as claimed. Therefore, Zn is cyclic. 4 we will see that Z and Zn (n ∈ N) are essentially the only cyclic groups in the sense that any cyclic group is isomorphic to one of these. 3 Example Prove that the group Q (under addition) is not cyclic. Solution We give a proof by contradiction. Suppose that Q is cyclic. Then Q = g for some g ∈ Q. Then g/2 ∈ Q = g , so g/2 = kg for some k ∈ Z. Now g = 0 (otherwise we get the contradiction Q = 0 = {0}), so we can divide both sides of g/2 = kg by g to get the contradiction 1/2 = k ∈ Z.

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Abstract Algebra I by Randall R. Holmes


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