Abstract Nonsense

Review of Group Theory: The Fourth Isomorphism Theorem (The Correspondence Theorem)

Point of post: In this post we describe the Fourth Isomorphism Theorem. We’ll then define simple groups and maximal subgroups and prove the fundamental result that $N\unlhd G$ is maximal if and only if $G/N$ is simple.

Motivation

Intuitively  the Fourth Isomorphism Theorem says that if $G$ and $G'$ are groups and $\phi\in\text{Hom}\left(G,G'\right)$ then there is an inclusion preservering bijective correspondence between the subgroups of $G$ contaning $\ker\phi$ and the subgroups of $G'$

The Fourth Isomorphism Theorem

We cut right to the chase:

Theorem (Fourth Isomorphism Theorem): Let $G$ and $G'$ be groups and $\phi:G \twoheadrightarrow G'$  (this means that $\phi$ is an epimorphism). Then, if $A=\left\{H\leqslant G:\ker\phi\leqslant H\right\}$ and $B=\left\{H'\leqslant G'\right\}$ then

$f:A\to B:H\mapsto \phi\left(H\right)$

is an order preserving bijection. Also, if $N\in A$ we have that $N\unlhd G$ if and only if $\phi\left(N\right)\unlhd G'$ and if this condition is satisfied then $G/N\cong G'/\phi(N)$.

Proof: To prove that $f$ is injective suppose that $H_1,H_2\in A$ and $\phi\left(H_1\right)=f\left(H_1\right)=f\left(H_2\right)=\phi\left(H_2\right)$. We note then that it must be true that $\phi^{-1}\left(\phi\left(H_1\right)\right)=\phi^{-1}\left(\phi\left(H_2\right)\right)$ but note that since $\ker\phi\subseteq H_i,\text{ }i=1,2$ we have that

\displaystyle \begin{aligned}\phi^{-1}\left(\phi\left(H_i\right)\right) &= \bigcup_{h\in H_i}\phi^{-1}\left(\phi(h)\right)\\ &= \bigcup_{h\in H_i}h\ker\phi\\ &= H_i\ker\phi\\ &= H_i\end{aligned}

from where it follows that $H_1=H_2$ and thus $f$ is injective. Now, to see that $f$ is surjective we let $H'\in B$. Note then that $\phi^{-1}\left(H'\right)\in A$ and $f\left(\phi^{-1}\left(H'\right)\right)=\phi\left(\phi^{-1}\left(H'\right)\right)=H'$ and thus $f$ is surjective, and so consequently bijective. To see that $f$ is inclusion preserving we merely note that $H_1\subseteq H_2$ then $f\left(H_1\right)=\phi\left(H_1\right)\subseteq \phi\left(H_2\right)=f\left(H_2\right)$.

To prove the second part of the theorem we merely note that if $N\unlhd G$ then $\phi\left(N\right)\unlhd G'$ (since $\phi$ is an epimorphism and we’ve already proven this). Conversely, suppose that $f\left(N\right)\unlhd G'$. Then, since $\displaystyle N\in A$ using the same techniques above we know that $\phi^{-1}\left(f\left(N\right)\right)=N$ and thus $N\unlhd G$ (since the preimage of normal subgroups are normal under homomorphisms as we’ve already proven). Now, assuming $N\unlhd G$ to prove that $G/N\cong G/\phi(N)$ we define

$\theta:G\to G/\phi(N):g\mapsto \phi(g)\phi(N)$

since $\phi$ is surjective clearly $\theta$ is surjective. Moreover, we see that since $N\in A$ we may use the techniques above to show that $\phi^{-1}\left(\phi\left(N\right)\right)=N$.  Thus,

\displaystyle \begin{aligned}\ker\theta &= \left\{g\in G:\theta(g)=\phi\left(N\right)\right\}\\ &= \left\{g\in G:\phi(g)\phi\left(N\right)=\phi\left(N\right)\right\}\\ &= \left\{g\in G:\phi(g)\in \phi\left(N\right)\right\}\\ &= \phi^{-1}\left(\phi\left(N\right)\right)\\ &= N\end{aligned}

Thus, by the First Isomorphism Theorem we may conclude that

$G/N\cong G'/\phi\left(N\right)$

the conclusion follows. $\blacksquare$

From this we have the following nice corollary:

Corollary: Let $G$ be a group and $N\unlhd G$. Then, for any $H'\leqslant G/N$ there exists a unique $N\leqslant H\leqslant G$ such that $H'=HN$. Moreover, $H\unlhd G$ if and only if $H'\unlhd G/N$.

Proof: Consider the canonical projection

$\pi:G\to G/N$

By the Fourth Isomorphism Theorem we have that the mapping

$f:\left\{H: N\leqslant H\leqslant G\right\}\to\left\{H'\leqslant G/N\right\}:H\mapsto \phi\left(H\right)$

is a bijection, consequently $H'=\phi\left(H\right)$ for some $H\in A$. But, $\phi(H)=HN$. The second part follows immediately from the second part of the Fourth Isomorphism Theorem. $\blacksquare$

Simple Groups and Maximal Subgroups

A group $G$ is called simple if $N\unlhd G$ implies that $N=\{e\}$ or $N=G$. In other words, a group is simple if it has no non-trivial normal subgroups. For a group $G$ a proper normal subgroup $N$ is called maximal if $H\subseteq K\unlhd G$ implies $K=H$ or $K=G$. We now use the Fourth Isomorphism Theorem to show how the two are realted:

Theorem: Let $G$ be a group and $N\lhd G$. Then, $N$ is maximal if and only if $G/H$ is simple.

Proof: This an immediate consequence of the last corollary to the Fourth Isomorphism Theorem. In particular, first assume that $N$ is maximal. Then, if $H'\unlhd G/N$ we have that $H'=HN$ for some $N\leqslant H\unlhd G$ but since $N$ is maximal this implies that $H=N,G$ so that $H'=N$ or $H'=G/N$ as required.

Conversely, suppose that $G/N$ is simple and $H\leqslant K\unlhd G$. Then, $KN\unlhd G$ and so $KN=GN$ or $KN=N$. But, by assumption this can only be true if $K=N$ or $K=G$ from where the conclusion follows. $\blacksquare$

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 4, 2011 -

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