# The Isomorphism Theorems

For a short explanation of this type of post, go here.

In this post I’m using Dummit and Foote [1] and Wikipedia for reference. As a minor disclaimer some of Dummit and Foote’s proofs are more general. I made a few extra assumptions that at least in my understanding still give us results that are sufficiently powerful for our purposes. As with all these theorems the proofs are simple. I have omitted large chunks, mostly the parts I consider verification, but if you’re following along I would recommend at least mentally checking off properties. Dummit and Foote actually introduce quotient groups using homomorphisms rather than defining normal subgroups and then quotient groups but while I think that’s a reasonable approach I don’t think it’s how most people first learned the concept. I went with the traditional approach of beginning with normal subgroups.

Anyway, let’s get on to the material. First a brief and common lemma will make my life a bit easier:

Lemma 1 When ${G}$ is a group and ${K \leq G}$, we have ${g_1K = g_2K \Longleftrightarrow g_2^{-1}g_1 \in K}$

Proof: If ${g_2^{-1}g_1 \in K}$ then ${g_2^{-1}g_1 = k}$ for some ${k \in K}$, and we have ${g_1 = g_2k}$, so ${g_1K = g_2kK = g_2K}$. Conversely if ${g_1K = g_2K}$ then for some ${k_1,k_2 \in K}$ we have ${g_1k_1 = g_2k_2}$ so ${g_1 = g_2k_2k_1^{-1} \Longrightarrow g_2^{-1}g_1 = k_2k_1^{-1} \in K}$. $\square$

As a last comment, in case this notation isn’t standard, Dummit and Foote use ${H \leq G}$ to say ${H}$ is a subgroup of ${G}$, ${N \trianglelefteq G}$ to say ${N}$ is a normal subgroup of ${G}$, and ${K \cong G}$ to say ${K}$ is isomorphic to ${G}$.

Theorem 2 (The First Isomorphism Theorem) Let ${\varphi: G \rightarrow H}$ be a surjective homomorphism. Then ${\ker(\varphi) \trianglelefteq G}$ and ${G / \ker(\varphi) \cong H}$. For ease of writing we’ll let ${K}$ be ${\ker(\varphi)}$.

Proof: First we’ll verify that ${\ker(\varphi) \trianglelefteq G}$.

• Closure: For ${k_1,k_2 \in K}$ we have ${\varphi(k_1k_2) = \varphi(k_1)\varphi(k_2) = e_He_H = e_H}$, so ${k_1k_2 \in K}$.
• Identity: Since for any ${g \in G}$ we have ${\varphi(g) = \varphi(e_Gg) = \varphi(e_G)\varphi(g)}$, ${\varphi(e_G) = e_H}$ and so ${e_H \in K}$.
• Inverses: For any ${k \in K}$ we have ${\varphi(k)\varphi(k^{-1}) = \varphi(kk^{-1}) = \varphi(e_G) = e_H}$, but since ${\varphi(k) = e_H}$ we also have that ${\varphi(k^{-1}) = e_H}$ and so ${k^{-1} \in K}$.

To show that ${G/K \cong H}$ we just need to show that ${\phi: G/K \rightarrow H}$ where ${\phi(gK) = \varphi(g)}$ is well-defined, a homomorphism, injective, and surjective.

• Well-defined: If ${g_1K = g_2K}$, then ${g_2^{-1}g_1 \in K}$ by lemma 1 and so ${\varphi(g_2^{-1}g_1) = e_H}$, ${\varphi(g_2^{-1})\varphi(g_1) = e_H}$, ${\varphi(g_2^{-1})\varphi(g_1) = e_H}$, and finally ${(\varphi(g_2)^{-1})^{-1} = \varphi(g_1)}$, so ${\varphi(g_2) = \varphi(g_1)}$ which means that ${\phi}$ is well defined.
• Homomorphism: Since ${\phi(g_1K)\phi(g_2K) = \varphi(g_1)\varphi(g_2) = \varphi(g_1g_2) = \phi(g_1g_2K) = \phi((g_1K)(g_2K))}$, ${\phi}$ is a homomorphism of groups.
• Injective: This amounts to reversing the proof that ${\phi}$ is well defined: We have ${\phi(g_1K) = \phi(g_2K) \Longrightarrow \varphi(g_1) = \varphi(g_2) \Longrightarrow \varphi(g_1)\varphi(g_2^{-1}) = e_H \Longrightarrow \varphi(g_1g_2^{-1}) = e_H \Longrightarrow g_1g_2^{-1} \in K}$, so we can conclude ${g_1K = g_2K}$ by lemma 1 and so ${\phi}$ is injective.
• Surjective: For any ${h \in H}$ there exists a ${g \in G}$ such that ${\varphi(g) = h}$ since ${\varphi}$ is surjective, and so ${\phi(gK) = \varphi(g) = h}$, so ${\phi}$ is surjective. $\square$

Dummit and Foote also point out an immediate consequence of this theorem that sort of sheds some light on its utility:

Corollary 3 If ${\varphi: G \rightarrow H}$ is a group homomorphism then ${\varphi}$ is injective iff ${\ker(\varphi) = e_G}$.

Proof: If ${\varphi}$ isn’t surjective then we need only consider its image, so we can just assume that it is. If ${\varphi}$ is injective then it’s a bijection so the statement is obvious. If ${\ker(\varphi) = e_G}$ then ${G / \ker(\varphi) = G}$ and so ${\varphi}$ is an isomorphism, so it’s injective. $\square$

A picture I like to maintain in my head looks like this:

Where ${K}$ is ${\ker(\varphi)}$, ${\varphi: G \rightarrow H}$, ${\phi: G/K \rightarrow H}$ is defined as ${\phi(gK) = \varphi(g)}$, and ${\pi: G \rightarrow G/K}$ is defined as ${\pi(g) = gK}$. We can think of ${\varphi}$ as ${\pi \circ \phi}$, and while this doesn’t immediately appear helpful, in the case where ${\ker(\varphi) = 1}$ we can see that this is actually a lot of information about ${\varphi}$, since it fixes ${G/K}$, in this case ${G}$. By relating ${\varphi}$ to this composition we preclude the idea that it might be mapping elements in an unequal fashion because of Lagrange’s theorem. In my mind I think of this as saying that the set of homomorphisms on ${G}$ are in some sense equivalent to the set of normal subgroups, although this is mostly speculation.

The next two theorems heavily utilize the first, so hopefully the result makes sense. We’ll see some application

Theorem 4 (The Second Isomorphism Theorem) Let ${G}$ be a group and let ${S,N \leq G}$ with ${N}$ normal. Then ${SN \leq G}$, ${S \cap N \trianglelefteq S}$, and ${(SN)/N \cong S/(S \cap N)}$.

Proof: The first two claims are a matter of verification, so I’ll skip those. As a result of those claims those we know that ${S \cap N}$ is a normal subgroup of ${S}$ so ${S/(S \cap N)}$ is well defined. For the other quotient group ${N}$ is normal in ${G}$ by assumption and since ${SN \leq G}$ it’s also normal in ${SN}$ (note that this is not guaranteed for some ${F}$ where ${G \leq F}$).

Here’s an outline of what we’re going to do. We want to show that ${S/(S \cap N) \cong (SN)/N}$, and we know that ${G/\ker(\varphi) \cong \varphi(G)}$. So it would be ideal if we could define a ${\varphi: S \rightarrow (SN)/N}$ such that ${\ker(\varphi) = S \cap N}$. If we do this then by the first isomorphism we’ll be done.

An obvious guess at how to define ${\varphi}$ is to let ${\varphi(s) = sN}$. Since all our quotient groups are legitimate this is well-defined and everything, and it’s easy to show that it’s a surjective homomorphism. Really all that remains to be shown is that ${\ker(\varphi) = S \cap N}$. The identity of ${(SN)/N}$ is ${N}$, so the kernel of ${\varphi}$ is the set of elements ${x \in S}$ where ${xN = N}$. By lemma 1 this implies that ${x \in N}$, and so the kernel of ${\varphi}$ is the set of elements in ${N}$ but also in ${S}$ by definition, so ${S \cap N}$. So we have a ${\varphi}$ such that ${\varphi: S \rightarrow (SN)/N}$ with ${\ker(\varphi) = S \cap N}$, so by the first isomorphism theorem we’re done. $\square$

I was actively trying to apply the first isomorphism theorem in that proof. I’m sure there are more direct ways to prove it, but I don’t think this one is too bad. It basically illustrates a situation where if you define a homomorphism you can make some non-obvious statements about the structure of ${G}$. In this case the second isomorphism theorem was here to tell me how to fit the pieces together to make the result seem nice, but the idea is more general. Here’s the relevant part of the lattice diagram of ${G}$, where quotient groups formed by subgroups joined by the red lines are isomorphic.

Dummit and Foote also call this the diamond isomorphism theorem. The third isomorphism theorem, stated next, is another example of showing isomorphism by defining the appropriate homomorphism.

Theorem 5 (The Third Isomorphism Theorem) Let ${G}$ be a group with ${H,K \trianglelefteq G}$ and ${H \leq K}$. Then ${(K/H) \trianglelefteq (G/H)}$ and ${(G/H)/(K/H) \cong G/K}$.

Proof: Again I’ll leave it to the reader to verify the first part. The last part is more fitting pieces into the first isomorphism theorem. Let ${\varphi: G/H \rightarrow K/H}$ be defined by ${\varphi(gH) = kH}$ which can be verified as a surjective homomorphism. Then ${\ker(\varphi)}$ is the set of cosets of the form ${gH}$ such that ${gK \in K}$, by again by lemma 1 this implies that ${g \in K}$, so the kernel is the set of cosets of the form ${kH}$, better known as the elements of ${K / H}$, and so by the first isomorphism theorem we’re done. $\square$

Putting aside the obvious jokes about canceling the ${H}$, after the first isomorphism theorem we pretty much went back to our symbol-manipulating ways. We could work through some examples of groups but I think there is only limited utility in that. A real application of the theorems is more desirable, and these do come up in solvable groups if I recall correctly. This gives me an excuse to get sidetracked in the Galois theory proof that quintics are not solvable in radicals. Personally this is one of my favorite results in all of math, which is why I was looking for an excuse to go learn about it.

Some authors, Dummit and Foote included, have a fourth isomorphism theorem that for a group ${G}$ and normal subgroup ${N}$ shows a bijection between the subgroups of ${G}$ which contain ${N}$ and the subgroups ${G/N}$. I don’t think this is any less useful than the other theorems, but I think enough algebra has been covered for our purposes. Next time I will write a little bit about definitions in topology, a subject I have no more than a week of formal education in, so despite the elementary level it will be a challenge. After that I will try to talk a little about Hilbert’s Nullstellensatz and basis theorem (which I am only marginally more educated in), and then we should be ready to at least attempt a foray into our end goal of algebraic geometry.

Sources:
[1] Dummit and Foote. Abstract Algebra, third edition, pages 97-99. Published 2004.
Used for the statement of isomorphism theorems.