1 EXAMPLE 9 Define phiZ_(12)rarrZ_(12) by phi(x)=3x . To verify that phi is a homomorphism, we chserve that in Z_(12),3(a+b)=3a+3b (since the group operation is addition modulo 12). Direct calculations show that Ker phi=(0,4,8) . Thus, we know from property 5 of Theorem 10.2 that phi is a 3*10-1 mapping. Since phi(2)=6 , we have by property 6 of Theorem 10.1 that phi^(-1)(6)=2+Ker phi=(2,6,10) . Notice also that (2) is cyclic and phi((2:))=(0,6) is cyclic. Moreover, 12 }=6 and |phi(2)|= |6|=2 , so ∣phi(2) divides |2| in agreement with property 3 of Theorem 10.1. Letting bar(K)=(0,6] , we see that the subgroup phi^(-1)( bar(K))=[0,2,4,6 , 8,10) . This verifies property 7 of Theorem 10.2 in this particular case. The next example illustrates how one can easily determine all homomorphisms from a cyclic group to a cyclic group. EXAMPLE 10 We determine all homomorphisms from Z_(12) to Z_(30) . By property 2 of Theorem 10.1, such a homomorphism is completely specified by the image of 1. That is, if 1 maps to a , then x maps to xa , Lagrange's Theorem and property 3 of Theorem 10.1 require that lai divide both 12 and 30 . So, |a|=1,2,3 , or 6 . Thas, a=0,15,10.20 , 5 , or 25 . This gives us a list of candidates for the homomorphisms. That each of these six possibilities yields an operation-preserving, welldefined function can now be verified by direct calculations. [Note that gcd(12,30)=6 . This is not a coincidence!]

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1 EXAMPLE 9 Define phiZ_(12)rarrZ_(12) by phi(x)=3x . To verify that phi is a homomorphism, we chserve that in Z_(12),3(a+b)=3a+3b (since the group operation is addition modulo 12). Direct calculations show that Ker phi=(0,4,8) . Thus, we know from property 5 of Theorem 10.2 that phi is a 3*10-1 mapping. Since phi(2)=6 , we have by property 6 of Theorem 10.1 that phi^(-1)(6)=2+Ker phi=(2,6,10) . Notice also that (2) is cyclic and phi((2:))=(0,6) is cyclic. Moreover, 12 }=6 and |phi(2)|= |6|=2 , so ∣phi(2) divides |2| in agreement with property 3 of Theorem 10.1. Letting bar(K)=(0,6] , we see that the subgroup phi^(-1)( bar(K))=[0,2,4,6 , 8,10) . This verifies property 7 of Theorem 10.2 in this particular case. The next example illustrates how one can easily determine all homomorphisms from a cyclic group to a cyclic group. EXAMPLE 10 We determine all homomorphisms from Z_(12) to Z_(30) . By property 2 of Theorem 10.1, such a homomorphism is completely specified by the image of 1. That is, if 1 maps to a , then x maps to xa , Lagrange's Theorem and property 3 of Theorem 10.1 require that lai divide both 12 and 30 . So, |a|=1,2,3 , or 6 . Thas, a=0,15,10.20 , 5 , or 25 . This gives us a list of candidates for the homomorphisms. That each of these six possibilities yields an operation-preserving, welldefined function can now be verified by direct calculations. [Note that gcd(12,30)=6 . This is not a coincidence!]

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We are given a series of mathematical statements that refer to group theory, specifically concerning homomorphisms from one group to another.

In the first part (Example 9), it is shown that a function ϕ: Z12 → Z12 sending x to 3x is a homomorphism. To demonstrate this, it is shown that 3(a + b) = 3a + 3b in Z12 where the group operation is addition modulo 12. The kernel of ϕ (Ker ϕ) is given as {0,4,8}, which indicates the elements of Z12 that ϕ sends to 0. Since ϕ(2) = 6, find the inverse image of 6 under ϕ, which is obtained by adding elements of the kernel to 2. This results in ϕ^-1(6) = {2, 6, 10}. Z12 is a cyclic group, and the image of cyclic group Z12 under ϕ is {0,6}, which is also cyclic. The order of 2 in Z12 is 6 and the order of ϕ(2) in Z12 is 2. As property 3 of Theorem 10.1 suggests, the order of ϕ(a) divides the order of a. Letting K={0,6} shows that the preimage of subgroup K under ϕ is {0, 2, 4, 6, 8, 10}, hence satisfying a property (indicated as '?') of Theorem 10.2.

In the second part (Example 10), the task is to find homomorphisms from Z12 to Z30. Property 2 of Theorem 10.1 indicates that a homomorphism is fully defined once the image of the generator 1 is determined. Images of 1 allowed by Lagrange's Theorem and Theorem 10.1, property 3, are those integers a that divide both 12 and 30 - yielding a = 1, 2, 3, or 6. Therefore, the possible images of 1 under these homomorphisms are 0, 15, 10, 20, 5, or 25, allowing for six possible homomorphisms. A further observation is made that gcd(12,30)=6, hinting at a non-coincidental relationship.

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