By Irena Swanson

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Gkl ) = e, which holds if and only if for all j = 1, . . , k, gjl = e. Thus |(g1 , g2 , . . , gk )| is a multiple of |gi | for all i, whence it is a multiple of lcm{|g1 |, . . , |gk |}. But it couldn’t be anything smaller, for otherwise some gil wouldn’t be e. 5 Let G and H be finite groups such that G ⊕ H is cyclic. Then G and H are cyclic and |G| and |H| are relatively prime. Proof. Suppose that G ⊕ H is generated by (a, b). Then G is generated by a, so G is cyclic. Similarly H = b is cyclic.

Why is this group well-defined? In other words, if aH = a′ H and bH = b′ H for some a, b, a′ , b′ ∈ G, why is abH = a′ b′ H? Work it out (and use that H is normal). Warning: When the group operation on G is +, the notation for the cosets is accordingly additive: a + H rather than aH. In particular, a coset in Z of a subgroup H is denoted a + H. A further warning against confusion: in Z, the subgroup generated by an integer m consists of elements of Z that are multiples of Z. ) as mZ. Thus here, mZ is a subgroup, and its cosets are subsets of the form a + mZ.

Then K ∩ H is a normal subgroup of K, and K/(K ∩ H) ∼ = KH/H. 6 Prove the second isomorphism theorem. 7 (Third Isomorphism Theorem) Let K and H be normal subgroups of G, and assume that K ⊆ H. Then H/K is a normal subgroup of G/K, and (G/K)/(H/K) ∼ = G/H. 8 Prove the third isomorphism theorem. 1 (Fundamental Theorem of Finite Abelian Groups) A finite commutative group G is isomorphic to nZ1 Z ⊕ nZ2 Z ⊕· · ·⊕ nZk Z for some positive integers n1 , n2 , · · · , nk . I’ll skip the proof now and return to it after ring theory, when we’ll have more elegant machinery then.