Define phi : Z12 times Z18 rightarrow Z6 by phi(([m]12, [n]18)) = [m + n]6. Prov

Question

Define phi : Z12 times Z18 rightarrow Z6 by phi(([m]12, [n]18)) = [m + n]6. Prove that phi is well defined. Prove that phi is a group homomorphism, and determine ker(phi). As a hint, I will tell you that ker(phi) is cyclic. Explicitly what does the first homomorphism theorem say about this situation?

Explanation / Answer

If m and n are relatively prime, we can explicitly display an isomorphism. Define f: Z --> Zn x Zm by f(k) = (k mod n, k mod m). It is easy to check that this yields a ring homomorphism. ker f = {k in Z : f(k) = (0 mod n, 0 mod m)} = {k in Z : n|k and m|k} = {k in Z : mn | k}, since m and n are relatively prime = (mn)Z. The only 'tricky' part is showing that f is surjective. Given (a, b) in Zn x Zm, we need to find k in Z such that f(k) = (a, b). Since m and n are relatively prime, there exist x, y in Z such that mx + ny = 1 Note that mx = 1 (mod n) and ny = 1 (mod m). Let k = a * (mx) + b * (ny). Then f(k) = (a * 1 + 0, 0 + b * 1) = (a, b) as required. ------ Finally, the first isomorphism theorem yields the desired result. Similarly we can show that Zm*Zn is isomorphic to Z(mn/gcd(m,n)) HenceZ12*Z18 isomorphic to Z36 To show that f is well defined, we need to show that whenever [a]36 = [b]36 in Z36, then f([a]36 ) = f([b]36) in Z6. If [a]36 = [b]36, then a - b = 36t for some t in Z. Thus a - b = 6(6t) and hence f([a]36) = [a]6 = [b]6 = f([b]36) as required. To show that f is a homomorphism, note that for any [a]36; [b]36 in Z36, f([a]36+[b]36) = f([a+b]36) = [a+b]6 = [a]6+[b]6 = f([a]36)+f([b]36) and f([a]36[b]36) = f([ab]36) = [ab]6 = [a]6[b]6 = f([a]36)f([b]36) as required.

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.