By James Wilson

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Therefore either k = 1 which implies a = 1 or ak−1 = 1 and aak−1 = ak−1 a = ak ≡ 1 (mod p) and so ak−1 = a−1 . Since Z∗p is closed to products, ak−1 ∈ Z∗p . So we may conclude our new set is closed to inverses and so it is a group under multiplication; Z∗p = Z× p – the largest group inside the multiplication. 3 Example: Suppose m is a composite positive integer (not 1), which means it is a multiple of some k, k = m. Since m is positive we may take k to be positive and it is less than m since it does not equal m.

A ∼ b implies a = b or a = b−1 . If a = b then b = a so b ∼ a. iv; so in general b ∼ a – symmetric. • a ∼ b and b ∼ c implies a = b or a = b−1 ; and b = c or b = c−1 . Naturally if a = b and b = c then a = c so a ∼ c. If a = b and instead b = c−1 then a = c−1 so a ∼ c. Next if a = b−1 and b = c clearly b−1 = c−1 so a ∼ c. iv; therefore, a ∼ c – transitive. So we have an equivalence relation on G and with it equivalence classes which partition the elements. iii we know inverses are unique and coupled with part iv of the theorem it is clear each equivalence class has at most 2 elements: [a] = {a, a−1 }.

So ψ defines the standard n-product for associative operators. 35 Hint(3/5): Use the given hint directly. Prove each mapping is well-defined and apply the theorem appropriately. 2 Homomorphisms and Subgroups 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Hint(2/5): For the counter example consider the a homomorphism between the multiplicative monoids of Z3 and Z6 . 1 Homomorphisms . . . . . Abelian Automorphism . . . Quaternions . . . . . . D4 in R2×2 . . . . . . . Subgroups .

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

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