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Isomorphism of relatively prime groups 
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#1
Mar812, 05:36 PM

P: 18

1. The problem statement, all variables and given/known data
Allow m,n to be two relatively prime integers. You must prove that Z(sub mn) ≈ Z(sub m) x Z(sub n) 2. Relevant equations if two groups form an isomorphism they must be onto, 11, and preserve the operation. 3. The attempt at a solution since m and n are relatively prime, the gcd(m, n) = 1. mhm, very stumped from the start. 


#2
Mar812, 07:50 PM

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What do you know about the group Z(sub mn)? What are some facts about it? Is it abelian, etc.?



#3
Mar812, 08:42 PM

P: 18

Isomorphisms isnt my thing, I do not know many facts about Z(sub mn). I can say the contents of Z(sub mn) are {[0],[1], [2],.... [mn]}. generated by a single element? Im not sure...Abelian? yes i believe so...



#4
Mar812, 09:11 PM

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Isomorphism of relatively prime groups
So if Z(sub mn) is isomorphic to Z(sub m) x Z(sub n), then that group would also have to be cyclic, right? Can you show that Z(sub m) x Z(sub n) is cyclic? 


#5
Mar912, 08:24 AM

P: 18

mhmm i believe so, but i dont know where to start since I am dealing with a cross product...



#6
Mar912, 09:42 AM

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If you had to guess at which element of Z_{m}xZ_{n} generated the whole group, what would it be?
If you're not sure, try some small examples like Z_{2}xZ_{3} or Z_{3}xZ_{4} and see what you can come up with 


#7
Mar912, 07:17 PM

P: 18

What would the elements of Z2 x Z3 look like again?



#8
Mar912, 08:27 PM

P: 85

Ordered pairs (a, b) such that a is from Z2, and b is from Z3.



#9
Mar912, 08:30 PM

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And if A and B are groups, then there's a natural way to define a group operation on A x B. Surely this is discussed in your textbook or lecture? There's no way you can proceed with this problem without first understanding what the group is. 


#10
Mar1012, 12:04 AM

P: 18

yes of course I just wanted to be refreshed and reassured...
Z2 x Z3 = {([0],[0]), ([0],[1]), ([0],[2]), ([1],[0]), ([1],[1]), ([1],[2])}... I cannot grasp how this could be generated by a single element though? 


#11
Mar1012, 12:40 AM

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#12
Mar1112, 12:03 PM

P: 18

the generator for the group Z(sub 2) x Z(sub 3) = <[1],[1]> and since the operation is addition, the generator for a group in the form Zn x Zm = <[1], [1]>. so Zn x Zm is cyclic and abelian.



#13
Mar1112, 08:03 PM

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So Z2 x Z3 is cyclic, and I assume you observed that it has order 6. That looks promising, because Z6 is also cyclic with order 6. Now you need to define an isomorphism [itex]\phi: Z_{6} \rightarrow Z_2 \times Z_3[/itex] between the groups. As a first step, I suggest choosing a generator [itex]g[/itex] for [itex]Z_6[/itex] and defining [itex]\phi(g)[/itex]. 


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