 By Randall R. Holmes

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Additional info for Abstract Algebra I

Example text

Hint: First handle the case n = 2. ) For the general case, use the fact that if A and B are 2 × 2 matrices, then A 0 0 I B 0 AB 0 = , 0 I 0 I where the first matrix on the left denotes the n × n matrix with A in the upper left corner and the (n − 2) × (n − 2) identity matrix I in the lower right corner and zeros elsewhere, and similarly for the other matrices. 4–3 Let G = R\{−1} (reals without −1). Define ∗ on G by x ∗ y = x + y + xy. 37 Prove that (G, ∗) is a group. ) 4–4 Let G be a group and let a, b ∈ G.

19 , 13 , 1, 3, 9, . . }. (b) In Z we have 3 = {m3 | m ∈ Z} (m3 is the additive notation of 3m ) = {. . , (−2)3, (−1)3, (0)3, (1)3, (2)3, . . } = {. . , −6, −3, 0, 3, 6, . . } = 3Z. 3 Theorem. If H is any subgroup of G with a ∈ H, then a ⊆ H. 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).

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. (b) We have 31 = 3, 32 = 9, 33 = 27, and in general, 3n = 1 for every positive integer n.