# Abstract Nonsense

## 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

January 2, 2011 -

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