Viewing <3> and <12> as subgroups of Z, prove that <3>/<12> is isomorphic to Z4. Similarly, prove that <8>/<48> is isomorphic to Z6. Generalize to arbitrary integers k and n.

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- Nov 7th 2009, 09:53 AM #1

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- Nov 7th 2009, 04:08 PM #2
For the first part <3> is the set of all integer multiples of 3, and is clearly a group under addition. And <12>, the set of all integer multiples of 12 is also clearly a group and a subgroup of <3>. Since addition is abelian the subgroup is normal, so that shows that <3>/<12> is a properly defined group. The four sets of all elements of the form 12n,12n+3,12x+6,12x+9 clearly have union <3> and are also equally clearly the cosets of <12> in <3>. So <12> has index 4 and <3>/<12> has order 4. Clearly <12>+3 has order 4 in the factor group, so there is an element of order 4 so it must be isomorphic to Z4. In fact the map that sends 12n+3k to 4n+k for k=0,1,2,3 is the required isomorphism.

It should be clear that we can easily generalise this argument to any pair of natural numbers a,b where a divides b.

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