# Abstract Nonsense

## 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 $\phi:G\to G'$ is a homomorphism then

$G/\ker\phi\cong\phi\left(G\right)$

Motivation

Having already discussed the idea of quotient structures it is natural to ask “How does $G/N$ relate to other objects?” The first such question might be: we know that if $\phi\in\text{Hom}\left(G,G'\right)$ then $\ker\phi\trianglelefteq G$. Thus, the quotient group $G/\ker\phi$ is well-defined. How exactly does this relate back to $\phi$? There must be some connection between this quotient group and the original $\phi$, and indeed there is. One can think of $\phi$ 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 $G/\ker\phi$ we have ‘identified’ all the problem spots with each other in the sense that in the resulting quotient group if $g$ and $g'$ are such that $\phi(g)=\phi(g')$ then $g\ker\phi=g'\ker phi$. 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 $G$ and $G'$ are groups and $\phi\in\text{Hom}\left(G,G'\right)$ then $\ker\phi\trianglelefteq G$. We claim first that in essence if $\phi(g)=\phi(g')$ then $g=g'k$ for some $k\in\ker\phi$. In essence, the things that map to $\phi(g)$ are $g$ times an “irrelevant factor” (in the sense that the image is the identity). Put more formally:

Theorem: Let $G$ and $G'$ be groups and $\phi\in\text{Hom}\left(G,G'\right)$. Then, for every $g\in G$ we have that

$\phi^{-1}\left(\phi(g)\right)=g\ker\phi$

(where the right hand side of the above a coset of $\ker\phi$).

Proof: It’s evident that $\phi^{-1}\left(\phi\left(g\right)\right)\supseteq g\ker\phi$ since

$\phi\left(g\ker\phi\right)=\phi(g)\phi\left(\ker\phi\right)=\phi(g)e=\phi(g)$

(where it should be noted that all of the above notation is loose notation for the product of subsets). Conversely, suppose that $g'\in\phi^{-1}\left(\phi(g)\right)$ then $g'=g\left(g^{-1}g'\right)$ and

$\phi\left(g^{-1}g'\right)=\phi(g)^{-1}\phi(g')=\phi(g)^{-1}\pi(g)=e_{G'}$

and thus $g^{-1}g'\in\ker\phi$. The conclusion follows. $\blacksquare$

With this we are now able to prove the theorem at hand namely the first isomorphism theorem:

Theorem (First Isomorphism Theorem): Let $G$ and $G'$ be groups and $\phi\in\text{Hom}\left(G,G'\right)$. Then,

$G/\ker\phi\cong\phi\left(G\right)$

Proof: Considering our prior theorem the map

$\Phi:G/\ker\phi\to G':g\ker\phi\mapsto \phi(g)$

is well-defined (in the sense if $g\ker\phi=g'\ker\phi$ then $\Phi(g)=\Phi(g')$). Moreover, we also see that the map is injective since if $\phi(g)=\phi(g')$ then $g'\in g\ker\phi$ and so $g'=gk$ for some $k\in\ker\phi$. Thus,

$g'\ker\phi=(gk)\ker\phi=g\ker\phi$

so that $\Phi$ is also injective. But, it’s also evident that $\Phi$ is surjective since if $\phi(g)\in\phi\left(G\right)$ then $\Phi\left(g\ker\phi\right)=\phi(g)$. Thus, it suffices to prove that $\Phi$ is a homomorphism. But, this follows since

$\Phi\left(\left(g\ker\phi\right)\left(g'\ker\phi\right)\right)=\phi(gg')=\phi(g)\phi(g')=\Phi\left(g\ker\phi\right)\Phi\left(g'\ker\phi\right)$

from where the conclusion follows. $\blacksquare$

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

January 2, 2011 -

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