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February 10th, 2018, 01:41 AM   #1
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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
Attached Images q4.jpg (14.6 KB, 4 views) Webp.net-resizeimage.jpg (83.6 KB, 9 views) February 10th, 2018, 07:30 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic 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 $a-b = 6t = 2 \times 3t$ for some integer $t$. So $2$ divides $a-b$ (i.e. $[a]_2 = [b]_2$) and $3$ divides $a-b$ (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. Tags isomorphism, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jim198810 Computer Science 0 May 9th, 2015 03:47 AM mathbalarka Abstract Algebra 2 November 4th, 2012 10:53 PM bewade123 Abstract Algebra 2 February 14th, 2012 04:12 PM mia6 Linear Algebra 1 November 10th, 2010 08:31 AM Ujjwal Linear Algebra 1 November 8th, 2008 05:48 PM

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