A Hungerford’s Algebra Solutions Manual by James Wilson

By James Wilson

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2 furthermore states that every infinite cyclic group is isomorphic (and thus equipollent) to the integers. Therefore |ab| = ℵ0 = |ba| even when infinite. When n = 0, (cac−1 )0 = e = cc−1 = cec−1 = ca0 c−1 . Suppose (cac−1 )n = n −1 ca c , then (cac−1 )n+1 = (cac−1 )n (cac−1 ) = can c−1 cac−1 = can+1 c−1 ; therefore, (cac−1 )n = can c−1 for all n ∈ N, by induction. Also, (cac−1 )−1 = (c−1 )−1 a−1 c−1 = ca−1 c−1 , which leads to (cac−1 )n = (cac−1 )−(−n) = ((cac−1 )−1 )n = (ca−1 c−1 )n = ca−(−n) c−1 = can c−1 , and so in general (cac−1 )n = can c−1 for all n ∈ Z.

D4 has only two elements of order 4: C and C 3 ; yet Q8 has six: A, −A, B, −B, AB and −AB; thus these groups are not isomorphic. Hint(3/5): For the forward direction show S is the equivalence class of the identity. 5 Subgroups. Let S be a nonempty subset of a group G and define a relation on G by a ∼ b if and only if ab−1 ∈ S. Show that ∼ is an equivalence relation if and only if S is a subgroup of G. Proof: (⇒) Suppose ∼ is an equivalence relation on G. S is nonempty and thus it contains an element a.

Then f (1) = (a, b). 1 shows (a, b) + (a, b) = (0, 0). Thus f (2) = (0, 0) = f (0). Therefore f is not injective. Thus f has no inverse mapping so no homomorphism f −1 exists such that f −1 f = 1Z4 . Therefore the groups are not isomorphic. erify the definioup. 11 41 Center. If G is a group, then C = {a ∈ G | ax = xa for all x ∈ G} is an abelian subgroup of G. C is called the center of G. Proof: The identity element has the property ex = x = xe for any element x ∈ G; thus e ∈ C so C is nonempty.

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