By Randall R. Holmes

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By the Subgroup Theorem, H is a subgroup of G. 3 Cyclic subgroup Let G be a group and let a be a fixed element of G. Put a = {am | m ∈ Z}. 1 Theorem. a is a subgroup of G. Proof. 2). ) We have e = a0 ∈ a . ) Let x, y ∈ a . Then x = am and y = an for some m, n ∈ Z. So xy = am an = am+n ∈ a . ) Let x ∈ a . Then x = am for some m ∈ Z. So x−1 = (am )−1 = a−m ∈ a . By the Subgroup Theorem, a is a subgroup of G. We call a the (cyclic) subgroup of G generated by a. 2 Example (a) Find 3 in Q× (under multiplication).

In particular, mn ≤ k. On the other hand k is the order of the cyclic subgroup of Zm ⊕ Zn generated by (1, 1), so k is less than or equal to the order of Zm ⊕ Zn , which is mn. We conclude that k = mn, so in fact (1, 1) = Zm ⊕ Zn . Therefore Zm ⊕ Zn is cyclic. 1, Zm ⊕ Zn ∼ = Zmn . 57 6 – Exercises 6–1 Let G be a cyclic group of order n. Prove that if d is a positive integer dividing n, then G has a subgroup of order d. 1. First work with the special case G = Z12 to get an idea for the proof. 1, which says the following: Every cyclic group is abelian.

We proceed by induction on r. If r = 1, then σ is the identity, which, according to our convention, is a product of transpositions with no factors. Assume that r > 1. We claim that (i1 , i2 )σ = σ , where σ = (i2 , i3 , . . , ir ). We have, (i1 , i2 )σ(i1 ) = i1 = σ (i1 ), (i1 , i2 )σ(ir ) = i2 = σ (ir ), (i1 , i2 )σ(ij ) = ij+1 = σ (ij ), (i1 , i2 )σ(k) = k = σ (k), 63 1 < j < r, k = i1 , i2 , . .

### Abstract Algebra I by Randall R. Holmes

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