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February 10th, 2018, 02:41 AM  #1 
Member Joined: Sep 2011 Posts: 83 Thanks: 1  Isomorphism problem
Hi, I have attached the question and the solutions to part a and b of this question. Would like someone to verify if I have done anything wrong. Greatly appreciate it! Thanks. Would also like to check if there is a simpler method to prove f is an isomorphism? Thanks 
February 10th, 2018, 08:30 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 119 Thanks: 38 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
In both parts, you've started your argument by assuming what you're trying to prove. In (a), you're being asked to prove that $[a]_6 = [b]_6 \implies ([a]_2, [a]_3) = ([b]_2, [b]_3)$, but you start by assuming this is true in the first place! What you could instead do is start with the idea in your second line to get something like: If $[a]_6 = [b]_6$ then $ab = 6t = 2 \times 3t$ for some integer $t$. So $2$ divides $ab$ (i.e. $[a]_2 = [b]_2$) and $3$ divides $ab$ (i.e. $[a]_3 = [b]_3$). Hence $f([a]_6) = ([a]_2, [a]_3) = ([b]_2, [b]_3) = f([b]_6)$. In part (b), you've got the same issue. You're being asked to prove $f$ is an isomorphism, but you start by assuming it's an isomorphism! It seems like you've done most of the work needed for the proof (computing $f([a]_6$ for each $a$, showing $f$ respects addition, though you are missing an argument to show it respects multiplication) but you've got your argument back to front. You might like to have another attempt, bearing all this in mind. 

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