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Z x Z/<(1,2)> isomorphic to Z
and
Z x Z x Z/<(1,1,1)> isomorphic to Z x Z
Thanks.
What have you done my friend? Without any context, I'm not sure how you would even want to proceed by this? The obvious thing to do, for the first example you want a mappingOriginally Posted by JSB1917
whose kernel is
. How about something like
, and
. Then,
which....
I have nothing really because I'm not sure what I'm supposed to exactly do.What have you done my friend? Without any context, I'm not sure how you would even want to proceed by this? The obvious thing to do, for the first example you want a mapping
Clearly you are going to want to construct an epimorphismwith
and then apply the FIT. Now, since
has basis
(i.e. it's free) you need only specify what you want your homomorphism to do to those guys, and then just extend by linearity. Now, how could you define a map on them, so that your kernel is what you want?
I have f(a,b)=2a-b for the first showed it's a homomorphism, showed the kern f = <(1,2)>, just need the onto.
Similarly, for the other one I have f(a,b,c)=(a-c,b-c), showed it's a homomorphism, kern f = <(1,1,1)>, just need the onto.
I just need help with onto.
pick a k in Z. now we need to find a pair of integers (a,b) such that f(a,b) = 2a-b = k. there could be a LOT of pairs with this property, we only need to find ONE. so let's make it easy on ourselves, and see if we can find a pair of the form (0,b) such that f(0,b) = - b = k.
well, it appears that b = -k, will do the trick. so
(we could have decided to let (a,b) = (a,k) (that is, choose b = k). then, f(a,k) = 2a - k = k implies that 2a = 2k, and we could choose a = k. and sure enough f(2k,k) = 2k-k = k. there's literally infinite ways we could have gone about them, and all are equally good. the trick is to pick "some" value for either a, or b, and then find out how we have to choose "the other part of the pair". for example, suppose we choose a = m. then f(a,b) = 2m - b. if 2m - b = k, then b has to be 2m - k. and sure enough, f(m,2m-k) = 2m - (2m - k) = 2m - 2m + k = k. see how that works?).
a similar logic applies to the second problem. we want to pick a,b,c so that a-c = k, b-c = m, for any arbitrary pair (k,m). one possible choice is (k,m,0). do you see the similarity here with solving a system of linear equations?
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