Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: The Second Isomorphism Theorem


Point of post: In this post we prove the second isomorphism theorem.

Motivation

As was stated in the post on the First Isomorphism Theorem there other isomorphism theorems which, using the first one, aren’t too hard to prove. The second one in essence says that it says that if you multiply two subgroups A,B  of a group G with B normal the resulting set  AB will be, in fact, a subgroup. Moreover, we find that AB/B\cong A/(A\cap B). In particular we get the interesting result that if A\cap B is trivial then AB/B\cong A, so you can “just cancel them!”

The Second Isomorphism Theorem

We begin by proving a few lemmas. Namely:

Lemma: Let G be a group with A\leqslant G and B\trianglelefteq G. Then, AB\leqslant G.

Proof: Recall that it suffices to prove that AB=BA.  Thus, we notice that

\displaystyle \begin{aligned}AB &= \bigcup_{a\in A}aB\\ &= \bigcup_{a\in A}Ba\\ &= BA\end{aligned}

\blacksquare

The next one is that

Lemma: Let G be a group, A\leqslant G and B\trianglelefteq G. Then, B\trianglelefteq AB.

Proof: We recall that since B\trianglelefteq G and AB\leqslant G that \left(B\cap AB\right)\trianglelefteq AB. Note though that B\subseteq AB and thus it follows that B\trianglelefteq AB. \blacksquare

We are now ready to state and prove our Second Isomorphism Theorem:

Theorem (Second Isomorphism Theorem): Let G be a group with A\leqslant G and B\trianglelefteq G then

\displaystyle AB/B\cong A/\left(A\cap B\right)

Proof: By our previous lemmas we know that this makes sense (in the sense that we can really consider the above quotient groups). Consider then the canonical projection

\pi:AB \to AB/B

Then, the restriction \pi_{\mid A} is a homomorphism

\pi_{\mid A}:A\to AB/B

Note though that \ker\left(\pi_{\mid A}\right)=A\cap B and

\displaystyle \begin{aligned}\pi_{\mid A}\left(A\right) &= \left\{aB:a\in A\right\}\\ &= \left\{(ab)B:a\in A\text{ and }b\in B\right\}\\ &= \left\{cB:c\in AB\right\}\\ &= AB/B\end{aligned}

Thus, by the First Isomorphism Theorem we may conclude that

A/\left(A\cap B\right)\cong AB/B

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

 

Advertisements

January 2, 2011 - Posted by | Algebra, Group Theory | , ,

3 Comments »

  1. […] . By first principles we know that . Moreover, we claim that is abelian for . Indeed, by the second isomorphism theorem we have […]

    Pingback by Review of Group Theory: Solvable Groups « Abstract Nonsense | March 11, 2011 | Reply

  2. […] so . We note then that since we have that , and since , and we  have by our lemma that . Similarly, one has that (recalling that […]

    Pingback by Relation Between Sylow’s Theorems and Direct Product « Abstract Nonsense | April 19, 2011 | Reply

  3. […] that . We recall then that since we have that and since so that is trivial we may conclude from a common […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2004) « Abstract Nonsense | May 6, 2011 | Reply


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 )

Twitter picture

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

Facebook photo

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

Google+ photo

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

Connecting to %s

%d bloggers like this: