By Randall R. Holmes

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Xn−1 } and the elements xi , 0 ≤ i < n, are all distinct. (iii) | x | = ord(x). Proof. (i) Assume that ord(x) = ∞. By definition, x consists of all powers xi with i ∈ Z, so the equation follows once we observe that x0 = e and x1 = x. ) Suppose that xi = xj with i ≤ j. Then xj−i = xj x−i = xj (xi )−1 = xj (xj )−1 = e. Now j − i is a nonnegative integer and, since ord(x) = ∞, this number cannot be positive. Therefore, j − i = 0, that is, j = i. This proves that the elements xi , i ∈ Z, are all distinct.

But R+ is not a subgroup of R since R+ does not use the same binary operation as R. The following theorem provides the most common way to check whether a subset of a group is a subgroup. 2 Subgroup Theorem. A subset H of the group G is a subgroup if and only if the following hold: (i) e ∈ H, G (ii) x, y ∈ H ⇒ xy ∈ H, H (iii) x ∈ H ⇒ x−1 ∈ H. y x e x−1 xy Note: (i) says that the identity element e of G is in H, (ii) says that H is closed under the binary operation of G, and (iii) says that H is closed under inversion.

If G has a property that can be described just using its elements and its binary operation, then G must have that same property, and vice versa. The next example illustrates this principle. 4 Example only if G is abelian. Assume that G ∼ = G . Prove that G is abelian if and Solution Since G ∼ = G , there exists an isomorphism ϕ : G → G . (⇒) Assume that G is abelian. 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.

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Abstract Algebra I by Randall R. Holmes

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