Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: The Third Isomorphism Theorem

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


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



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 - Posted by | Algebra, Group Theory | , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: