March 28, 2017 | | By admin | By Irena Swanson

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Extra resources for Abstract Algebra [Lecture notes]

Sample text

Observe that HK is not a subgroup here! 2 Let H and K be subgroups of G, and assume that H is normal in G. Then HK is a subgroup of G. Proof. Let h, h′ ∈ H, k, k ′ ∈ K. Then (hk)(h′ k ′ )−1 = hkk ′−1 h′−1 ∈ Hkk ′−1 H = H(Hkk ′−1 ) ⊆ HK. 3 Let G be a group with normal subgroups H and K such that H ∩ K = {e} and HK = G. Then G ∼ = H ⊕ K. Proof. Each g ∈ G can be expressed as g = hk for some h ∈ H and k ∈ K. These g and h are unique, for if hk = h′ k ′ for some h′ ∈ H and k ′ ∈ K, then (h′ )−1 h = k ′ k −1 ∈ H ∩ K = {e}, whence h′ = h and k ′ = k.

I’ll skip the proof now and return to it after ring theory, when we’ll have more elegant machinery then. 3. We can check out an example now. For example, we start with the group Z⊕Z⊕Z, the shorthand for which is Z3 . This is certainly a commutative group. Thus every subgroup is normal, so we will take Z3 G= . (1, −1, 1), (5, 1, −5), (−3, −3, 29) Then G is certainly a commutative group. It is not clear at all that it is finite. We will rewrite G with different generators, and for this we will suggestively record the relation vectors as columns in a matrix:   1 5 −3  −1 1 −3  .

5 Prove that the set Z[i] = {a + bi : a, b, ∈ Z} is a subring of C. The ring Z[i] is called the ring of Gaussian integers. Find, with proof, all the units in Z[i]. 6 Find all √ the units in Z[ 2]. Two of the units are 1 and 1 + 2. Let √ a + b 2 be a unit in Z[ 2], where √ a, b are integers. We want to find all a, b. Then √ there exist c, d ∈ Z such that (a + b 2)(c + d √2) = 1. This means that ac + 2bd = 1 √ and ad + bc = 0. Necessarily then also (a − b 2)(c − d 2) = 1, and multiplying the 2 2 2 2 four terms gives√us (a2 − 2b2 )(c √ − 2d ) = 1.