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Subgroups . . . . . . . Finite subgroups . . . . . nZ . . . . . . . . . Subgroups of Sn . . . . . Subgroups and Homomorphisms Z2 ⊕ Z2 lattice . . . . . . Center . . . . . . . . Generators . . . . . . . Cyclic Images . . . . . . Cyclic Groups of Order 4 . . . Automorphisms of Zn . . . . Generators of PruferGroup . . Join of Abelian Groups . . . Join of Groups . . . . . . Subgroup Lattices . . . . . . . . . . . . . . .

1 part (iv) which is where abelian comes into play) we see that a + b is a torsion element, a + b ∈ T , so T is closed. 3 we know |a| = | − a|, so T is closed to inverses. 5. 2. 10 Infinite Cyclic Groups. An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups. 2 it is isomorphic to Z. 2 shows the groups mZ are subgroups of Z which are furthermore isomorphic to Z. Consider any subgroup H = 0 of Z. 1 we know H = m = {mk | k ∈ Z} = mZ, where m is the least positive integer in H.

Be elements of an abelian group G such that |x1 | = p, px2 = x1 , px3 = x2 , . . , pxn+1 = xn , . . The subgroup generated by the xi (i ≥ 1) is isomorphic to Z(p∞ ). ] Proof: (a) Every element in Z(p∞ ) is of the form a/pi for some i ∈ N. In taking the sum of a/pi , pi times we have a/pi +· · ·+a/pi = pi (a/pi ) = a ∼ 0; therefore, (pi )a/pi = 0. 6, (Euclid’s Lemma) it must be pn for some n ≥ 0. (b) Let H be a subgroup of G with an element a/pi of highest order, pk , in H. i ) i Note in Q that a/pi = pa/(a,p i /(a,pi ) which is the reduced fraction ( (ai , p ) = 0 since pi = 0).

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