## 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 of a group with normal the resulting set will be, in fact, a subgroup. Moreover, we find that . In particular we get the interesting result that if is trivial then , so you can “just cancel them!”

*The Second Isomorphism Theorem*

We begin by proving a few lemmas. Namely:

**Lemma: ***Let be a group with and . Then, .*

**Proof: **Recall that it suffices to prove that . Thus, we notice that

The next one is that

**Lemma:*** Let be a group, and . Then, .*

**Proof: **We recall that since and that . Note though that and thus it follows that .

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

**Theorem (Second Isomorphism Theorem): ***Let be a group with and then*

**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

Then, the restriction is a homomorphism

Note though that and

Thus, by the First Isomorphism Theorem we may conclude that

as desired.

**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

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