# Abstract Nonsense

## Review of Group Theory: The Third Isomorphism Theorem

Point of post: In this post we discuss the Third Isomorphism Theorem.

Motivation

The third isomorphism theorem has to deal with the situation when $K, H\trianglelefteq G$  with $K\subseteq H$. It asks us if the simple arithmetic fact $\displaystyle \frac{\frac{a}{b}}{\frac{c}{b}}=\frac{a}{c}$ somehow applies to groups. In essence, it asks if $\left(G/K\right)/\left(H/ K\right)\cong G/H$?

The Third Isomorphism Theorem

We begin with a lemma:

Lemma: Let $H,K\trianglelefteq G$ with $K\subseteq H$. Then, $\left(H/K\right)\trianglelefteq \left(G/K\right)$.

Proof: Merely note that the canonical projection

$\pi:G\to G/K$

is a surjective homomorphism, $H\trianglelefteq G$, and $H/K=\pi\left(H\right)$. The result follows from an earlier theorem. $\blacksquare$

Now, the rest is easy:

Theorem (Third Isomorphism Theorem): Let $G$ be a group with $H,K\trianglelefteq G$ and $K\subseteq H$. Then,

$\left(G/K\right)/\left(H/K\right)\cong G/H$

Proof: This is simple as noticing that

$\phi:G/K\to G/H:gK\mapsto gH$

is evidently a surjective homomorphism and

\displaystyle \begin{aligned}\ker\phi &= \left\{gK\in G/K:gH=H\right\}\\ &= \left\{gK\in G/K:g\in H\right\}\\ &= H/K\end{aligned}

Thus, by the First Isomorphism Theorem we may conclude that

$\left(G/K\right)/\left(H/K\right)\cong G/H$

as desired. $\blacksquare$

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 -