Review of Group Theory: The First Isomorphism Theorem
Point of post: In this post we prove the first isomorphism theorem which, in essence says that for any homomorphism the image is isomorphic to the domain with a small perturbation. Precisely what this perturbation turns out to be is the kernel of the homomorphism. Explicitly we prove that if is a homomorphism then
Having already discussed the idea of quotient structures it is natural to ask “How does relate to other objects?” The first such question might be: we know that if then . Thus, the quotient group is well-defined. How exactly does this relate back to ? There must be some connection between this quotient group and the original , and indeed there is. One can think of as an “almost isomorphism” in the sense that it’s an isomorphism satisfying the temporary issue of non-injectivity. Consequently, one may ask “is there a way to ‘throw out’ the problem elements”? This is a familiar ideology to those working in analysis where one ‘mods out’ by violators of the positive semi-definitness of a metric or norm. It turns out that, for all intents and purposes, the answer is yes. Moreover, the way one does it is kind of what ‘seems natural’, especially if one is familiar with general topology or the above ideas of analysis (or more generally the set-theortic notion of kernel). In essence, we’ll see that by considering we have ‘identified’ all the problem spots with each other in the sense that in the resulting quotient group if and are such that then . Thus, the resulting space will be one for which there is a canonical ‘reduction’ of the original surjective homomorphism but one for which the ‘disease’ of injectivity is ‘cured’.
The First Isomorphism Theorem
Recall from our last post that if and are groups and then . We claim first that in essence if then for some . In essence, the things that map to are times an “irrelevant factor” (in the sense that the image is the identity). Put more formally:
Theorem: Let and be groups and . Then, for every we have that
(where the right hand side of the above a coset of ).
Proof: It’s evident that since
(where it should be noted that all of the above notation is loose notation for the product of subsets). Conversely, suppose that then and
and thus . The conclusion follows.
With this we are now able to prove the theorem at hand namely the first isomorphism theorem:
Theorem (First Isomorphism Theorem): Let and be groups and . Then,
Proof: Considering our prior theorem the map
is well-defined (in the sense if then ). Moreover, we also see that the map is injective since if then and so for some . Thus,
so that is also injective. But, it’s also evident that is surjective since if then . Thus, it suffices to prove that is a homomorphism. But, this follows since
from where the conclusion follows.
1. Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.
2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200