Abstract Nonsense

Crushing one theorem at a time

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.


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


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 - Posted by | Algebra, Group Theory, Uncategorized | , , , , ,


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