By Randall R. Holmes

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**Extra resources for Abstract Algebra I**

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Proof. Let H ≤ G with a ∈ H. We have, using closure, a1 = a ∈ H, a2 = aa ∈ H, a3 = a2 a ∈ H and in general am ∈ H for every positive integer m. Also, for each positive integer m, we have, using closure under inversion, a−m = (am )−1 ∈ H (since am ∈ H as was just shown). 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) = ∞.

2)3, (−1)3, (0)3, (1)3, (2)3, . . } = 3, so 3Z itself is cyclic. The following theorem shows that this is no coincidence. 1 Theorem. Every subgroup of a cyclic group is cyclic. Proof. Let G be a cyclic group and let H be a subgroup of G. Since G is cyclic, there exists g ∈ G such that G = g . If H = {e}, then H = e and H is cyclic. Now suppose H = {e}. Since every element of G, and hence H, is a power of g, it follows that g n ∈ H for some nonzero integer n. Now g −n = (g n )−1 ∈ H as well, so by replacing n by −n, if necessary, we may assume that n is positive.

Hint: From linear algebra, we know that a square matrix is invertible if and only if its determinant is nonzero. Use the fact that det(AB) = det(A) det(B) for A, B ∈ Matn (R). 5–3 (a) Find the order of the element 9 in Z15 . (b) Find the order of the matrix A = 0 −1 in the group Mat2×2 (R). 1 0 (c) Find the order of the matrix A = 0 −1 in the group GL2 (R). 1 0 5–4 (a) Compute 6 in the group Z15 . (b) Compute σ , where σ is the element of S4 given by σ= 1 2 3 4 . 3 1 4 2 51 5–5 Compute A in the group GL2 (R), where A = 1 1 .