Abstract Algebra

[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

[humph]

Hah, I remember this question. It's quite obvious if you don't overcomplicate it: just take and any subgroup (as is abelian, every subgroup is normal in ).

[/0]

Oh right, yeah does work. I was trying to do it for for some reason , but I think (at least) most subgroups of are isomorphic to that, because for all you can take , , but for you don't get a surjective map from . Bah, in the end I said no example was possible lol. If only I tried the most obvious quotient group

[/0]

When R is a commutative ring, and N is the set of nilpotents elements of R, prove that the only nilpotent element of R/N is zero.







But if how can I show that ?

After all, in ,