By James Wilson

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**Extra resources for A Hungerford’s Algebra Solutions Manual **

**Sample text**

Certainly f |A : A → f (A) is a well-defined function (refer to Introduction, section 3). Now f (ab) = f (a)f (b) for all elements in G so it must also for all element in A, which naturally come from G. Therefore f |A is a homomorphism. 2) f (A) is a group and so it is a subgroup of H. Proof: Hint(1/5): Show any homomorphism Z4 → Z2 ⊕ Z2 cannot be injective. 10 Z2 ⊕ Z2 lattice. List all subgroups of Z2 ⊕ Z2 . Is Z2 ⊕ Z2 isomorphic to Z4 ? Example: Z2 ⊕ Z2 (1, 0) (1, 1) 0 Z4 (0, 1) 2 0 Proof: Suppose f : Z4 → Z2 ⊕ Z2 is a homomorphism.

36 37 37 37 38 38 39 39 40 40 40 41 41 42 42 44 44 44 45 Homomorphisms. If f : G → H is a homomorphism of groups, then f (eG ) = eH and f (a−1 ) = f (a)−1 for all a ∈ G. Show by example that the first conclusion may be false if G, H are monoids that are note groups. Proof: Assuming f : G → H is a homomorphism of groups, then f (a) = f (aeG ) = f (a)f (eG ) and likewise on the left, f (a) = f (eG a) = f (eG )f (a).

D) The only proper subgroups of Z(p∞ ) are the finite cyclic groups Cn = 1/pn (n = 1, 2, . . ). Furthermore, 0 = C0 ≤ C1 ≤ C2 ≤ C3 ≤ · · · . 2 we proved a/pi = a1/pi . In part (e) notice G is defined in terms of equivalence classes so the case for well-defined must be explicit. Groups 50 (e) Let x1 , x2 , . . 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.

### A Hungerford’s Algebra Solutions Manual by James Wilson

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