Try mapping Z to Z with the map n -> mn, for some fixed m. Then let N=kmZ, k>1.
I am trying to construct an example of a theorem from my Algebra text.
At this point all I am trying to do is come up with a non-trivial example of the requirements of the theorem, that is to say a group N such that N is not simply the kernel of f. I started with D4 but can't think of a good group to make a homomorphism with. I might not be thinking big enough...If is a homomorphism of groups and N is a normal subgroup of G contained in the kernel of f, then there is...
Thanks!
-Dan
you can use D4, it actually provides a good example. in this case, let H be a cyclic group of order 2. the idea is we map all the rotations to e in H, and all the reflections to x ≠ e in H.
now the kernel of such a homomorphism is the rotations {1,r,r^2,r^3}, and it turns out that we can take N to be the normal subgroup of D4: {1, r^2}.
(the normality of this can be seen from the fact that this is the center of D4 (i think in some other post you called "r" C1).).
the fact that f: D4-->H is a homomorphism can be summed up geometrically as "a reflection followed by another reflection is a rotation".