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Abstract Algebra

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kamil9876:
It's really the Chineese remainder theorem. ie let us prove it for and .

We know that for every and there exists a unique solution such that:





So you can define your isomorphism as just . You know it's a (well-defined) bijection, now just check that it is a homomorphism.

/0:
Ah right, that makes sense, thanks kamil

humph:
Look up a proof! This was on one of my Algebra 1 assignments back in the day, actually. The trick is to consider generators from each group and show that the product of the two generators generates the product group.

kamil9876:
One of my favourite Paul Halmos quotes:

"For Heaven's sake don't look it up in a book, looking it up in a book is giving up. ..."

/0:
A question on the exam I wasn't sure about:

Find an example of a quotient group , where is a normal subgroup, which is not isomorphic to a subgroup of . Or prove that such an example doesn't exist.


Thanks

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