I suppose I'd like to focus on the theorems for groups and rings, first of all. In particular, I'd rather not see anything about modules, simply because I don't feel I know enough about them.

Anyway, in doing a lot of exercises in group theory and ring theory, I find that I use the first and third isomorphism theorems a lot more than the second and fourth. What I want to see are some really useful-looking applications of the second and fourth theorems that I may not have been exposed to before. Thanks!

EDIT: for clarity, the theorems I identify as the second and fourth are as follows for rings (cf. Dummit & Foote, 3ed p. 246)

The second: Let $A$ be a subring of $R$ and $B$ be an ideal of $R$. Then $A+B = \{a+b|a \in a, b \in B\}$ is a subring of $R$, $A \cap B$ is an ideal of $A$, and $(A+B)/B \cong A/(A \cap B)$.

The fourth: Let $I$ be an ideal of $R$. The correspondence $A \leftrightarrow A/I$ is an inclusion preserving bijection between the set of subrings $A$ of $R$ that contain $I$ and the set of subrings of $R/I$. Furthermore, $A$ (a subring containing $I$) is an ideal of $R$ if and only if $A/I$ is an ideal of $R/I$.

  • $\begingroup$ It might help to tell us which ones are the 2nd and 4th. I think the naming depends on the source. $\endgroup$ – user45861 Jun 25 '14 at 23:14
  • $\begingroup$ True enough. I'll edit my question now. $\endgroup$ – Samuel Yusim Jun 25 '14 at 23:17
  • $\begingroup$ possible duplicate $\endgroup$ – user17794 Jun 25 '14 at 23:17
  • $\begingroup$ Not entirely. That guy's second is my third. Give me a second to edit my post. $\endgroup$ – Samuel Yusim Jun 25 '14 at 23:19
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    $\begingroup$ The use of the 4th Iso Thm. in proving the Sylow Existence Theorem is pretty smooth I think. $\endgroup$ – Forever Mozart Jun 25 '14 at 23:30

This answer includes an application of the 4th isomorphism theorem for rings, and one (well, kinda two) of the 2nd isomorphism theorem for groups. I also link to a really cool application of the 4th isomorphism theorem in groups (but is kinda the same as the application to rings).

Rings, 4th isomorphism theorem

An neat application of the 4th isomorphism theorem for rings is the following:

Theorem: Maximal ideals are prime.

Proof: Let $R$ be a ring and let $I$ be a maximal ideal of $R$. Then consider the quotient $R/I$, and apply the correspondence theorem.

Indeed, if $R$ is commutative with identity then $R/I$ is a field. See here for a proof.

Groups, 4th isomorphism theorem

There is a rather cool trick in the theory of infinite groups, which was used by Higman to construct an infinite simple group. The idea is to appeal to Zorn's lemma and obtain a maximal normal subgroup, and then quotient this out to get a simple group. See here for a neat application of it. This is very similar to the above application to rings. (You quotient out a maximal subgroup/ideal to get a simple group/prime ring.)

Groups, 2nd isomorphism theorem

A nice application of the 2nd isomorphism theorem for groups is the following. It deals with soluble groups, which are an important class of groups (and are often taught in a second course on groups). A group $G$ is soluble (or solvable) if it possesses an abelian series, that is, a series $$1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$$ where each factor $G_{i+1}/G_i$ is abelian.

Theorem: Subgroups and homomorphic images of soluble groups are soluble.

Proof: Suppose $G$ is soluble with abelian series $1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$.

Subgroups: If $H$ is a subgroup of $G$ then we can apply the 2nd isomorphism theorem to get the following. $$\frac{H\cap G_{i+1}}{H\cap G_i}\cong\frac{(H\cap G_{i+1})G_i}{G_i}\leq\frac{G_{i+1}}{G_i}$$ Hence, the groups $\frac{H\cap G_{i+1}}{H\cap G_i}$ are abelian so the set $\{H\cap G_i; i=0, 1, \ldots, n\}$ forms an abelian series for the group $H$, as required.

Homomorphic images: If $N$ is a normal subgroup of $G$ then the can apply the 2nd isomorphism theorem to get the following. $$\frac{G_{i+1}N}{G_iN}\cong\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$$ The subgroup $G_i$ is a subgroup of $G_{i+1}$ and of $G_iN$, and so $G_i\lhd G_{i+1}\cap (G_iN)$. Therefore, $\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$ is a homomorphic image of $\frac{G_{i+1}}{G_i}$, and hence is abelian. By the 3rd isomorphism theorem we have the following. $$\frac{G_{i+1}N/N}{G_iN/N}\cong \frac{G_{i+1}N}{G_iN}$$ Hence, the set $\{G_iN/N; i=0, 1, \ldots, n\}$ forms an abelian series for $G/N$, as required.

  • $\begingroup$ All the answers I've gotten are great, but this one's really something. I appreciate it! $\endgroup$ – Samuel Yusim Jun 27 '14 at 0:36
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    $\begingroup$ Would the downvoter care to comment?... $\endgroup$ – user1729 Jun 27 '14 at 8:13

Notice as a special case $(M+N)/N=M$ if the sum is direct (meaning $M\cap N=0$). This is frequently used. For instance if $X$ and $Y$ are sets, and $FX$ is the free group on $X$ we have $F(X\coprod Y)=FX\oplus FY$, so that $F(X\coprod Y)/FY=FX$ is free, this is sometimes useful in algebraic topology.

As for the fourth: let me assure you that if you study a bit of commutative algebra and algebraic geometry you use this ALL THE TIME, its basically useful to figure out the ideals of quotient rings, which come up there.


The fourth isomorphism theorem, as you call it (I call it the correspondence theorem), is used alot in Algebraic Geometry. A good example of how it is used is to show that there is a homeomorphism between the topological spaces $$\text{Spec}(A)\simeq \text{Spec}(A/\sqrt{0})$$

Where $A$ is a commutative ring and $\sqrt{0}$ is its nilradical. If you are not familiar with $\text{Spec}(A)$, as a set it is the set of prime ideals in $A$. One can then put a topology on this set by declaring the closed sets to be of the form $$ V(I) = \{p\in\text{Spec}(A) : I\subset p\}.$$

Here $I$ is an ideal in $A$.

The correspondence theorem allows us to check that $\text{Spec}(A)$ and $\text{Spec}(A/\sqrt{0})$ are in bijective correspondence with each other. Indeed, it says that prime ideals in $A$ are in bijection with prime ideals in $A/\sqrt{0}$ that contain $\sqrt{0}$. However, every prime ideal contains $\sqrt{0}$ thus giving the bijection.

It then required more work to show that two topological spaces are homeomorphic, but as far I can recall, that does not require the correspondence theorem.


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