# Understanding the isomorphism theorems

1. Jun 1, 2015

### Avatrin

I am currently trying to understand the isomorphism theorems. The issue I am having is that I am struggling to find a way to think about them.

In Stillwell's Elements of Algebra, I found a way to understand the first theorem ($\frac{G}{ker \phi} \simeq Im \phi$ for any homomorphism $\phi:G\rightarrow G'$). It was proven in terms of set functions (since there is a one-to-one correspondence between the elements $e \in Im \phi$ and $\phi^{-1}(e)$). However, the second and third theorems are not even shown. Also, my curriculum is taken from Fraleigh's A First Course in Abstract Algebra. There, the first isomorphism theorem is longer than just the isomorphism between $\frac{G}{ker \phi}$ and $Im\phi$; He adds that there is a unique isomorphism $\mu: G/ker\phi \rightarrow \phi[G]$ such that $\phi(x) = \mu(\gamma(x))$ for each $x \in G$. Here, $\gamma$ is the canonical homomorphism from G to $\frac{G}{ker\phi}$.

I still do not understand what the canonical homomorphism is since it is not in the index of the book. More importantly, I am really having a hard time just thinking about the definitions involved. I can see a proof, and understand why each step is accurate. Yet, it is too abstract for me to process.

This has happened once before when I was learning measure- and integration theory. I couldn't understand anything for months until one day I read a definition of the Lebesgue integral in Euclidean space, and suddenly everything just "clicked"; Within a week I could understand, rather than just know, the entire curriculum. I am hoping for another one of those moments before exam.

I have the same problem now with certain aspects of abstract algebra. They generally involve factor groups and/or mappings. So, I have a suspicion that if I "get" the isomorphism theorems, the rest will connect.

2. Jun 1, 2015

### Orodruin

Staff Emeritus
It is just the homomorphism which arises by mapping every element in $G$ to its equivalence class.

3. Jun 2, 2015

### mathwonk

i.e. it takes x to [x]. you can't get much more canonical than that. think of the mapping that takes each student to the school he attends. you can identify all the students that attend the same school into one subset of all students. I.e. two students are equivalent if they attend the same school. Then there is one equivalence class of students for each school. that sets up a canonical correspomndence between the collection of all schools and the collection of all equivalence classes of students. I.e. in a sense there is no difference betwen a school, one thing, and the collection of all students at that school, a collection of things. The theory of sets is a technique for considering a collection of things to be one thing, a set. that's all that is going on in these theorems. the collection f^-1(y) of elements that map to the same image element y is thought of as representing that image element.

4. Jun 2, 2015

### Avatrin

Mathwonk: Except, my point in the original post was that the set theoretical language of Stillwell's book made me understand that theorem. The set theory is not the problem. When we mix some abstract algebra into it, is when I get perplexed. In most algebra books I have read, the books almost ignore that we are working on sets. Instead, algebraic concepts like factor groups and homomorphisms are used. Of course, a group is a set, but there are more rules involved. That complicates things a lot.

The main point in my original post was that I am having a hard time thinking this way.

Also, Stillwell only proves one out of three theorems.

5. Jun 2, 2015

### mathwonk

ok, yes it is true that the algebra adds more complexity, but the set theory is the irst step. second step is to show that when you have a operation, it is preserved by the equivalence relation. i.e. if you want to add two elements like A and B, getting A+B, and you change the elements to equiovalent ones, like A' and B', then you must check that A'+B' is equiovalent to A+B. If this is true then you can add classes, just by choosing any reprsentative elements, adding those, and passing to their class.

E.g. if we formthe quotient group of the integers by the (normal) subgroup of even integers, we get the quotient group {E,O}, where E represents all even integers and O represents all odd integers. Here if we add any two even integers we always get and even integer, i.e. if we choose any two integers representing zero, we always get an integer representing zero, and if we add an even and an odd integer we always get an odd integer. also two odds always add to an odd. so it makes sense to say that E+E = E, O+E = O, and O+O = E. This is a group with identity element E.

A group ios just a set plus an operation, so after verifying the set theorwtic part, you then just have to check the operation respects the separation into subsets. It actually makes many things easier than when they are just sets. E.g. in a quotient group, each equivalence class (here called a coset) has the same number of elements as every other coset.

One nice way is to think in terms of examples, like the group of rotations acting on the vertices of a cube. Then there is a subgroup leaving fixed a given vertex. and its cosets are the subsets of elements that take that vertex to the same vertex. I.e. two elements are equivalent if and only if they take that given vertex to the same resulting vertex. this subgroup is not normal, so the collection of cosets does not form a group in this case. you can try to check that composition of two equivalent elements does not always give equivalent results.

6. Jun 3, 2015

### jackmell

I'm studying Abstract Algebra too and had this problem as well. Thought I'd never understand the concept, but I found plowing-through the (group and ring) isomorphism problems over several weeks solved my problem (at least at a basic level). But I did more than just work the problems, rather I investigated the answers with real examples, actually programming in Mathematica the problems and seeing directly and concretely the answers develop. Then I experimented with the problems. Then observed how that changed the results. Dig, probe, experiment. Ask "why is that?". Then "what happens if I try changing it to something else?" Then change it and observed what happened. If you do this for a few weeks, then I bet you'll get it too.

If you're interested, here's a proof of the second isomorphism theorem for rings:

Also, here's an interesting concept I think you should understand about the isomorphism theorems: what are the consequences of the theorems when the homomorphism is onto and when it is not onto?

Here's something else I believe is helpful: Try and write a description of the following diagrams and then actually program in Mathematica examples illustrating the second one:

Last edited: Jun 3, 2015
7. Jun 3, 2015

### Avatrin

Well, I don't really need to learn the isomorphism theorems for rings. Also, Mathematica is not available here.

When the homomorphism is onto, $\frac{G}{ker\phi} \equiv G'$ since $Im(\phi) = G'$. If the homomorphism is not onto, $\frac{G}{ker\phi}$ is only isomorphic to a subset of G'.

8. Jun 3, 2015

### micromass

When I started learning abstract algebra, I had the same problem with this subject. I understood the formalism, but I could not understand the isomorphism theorems. I felt there was something missing. Then I found out later that there wasn't anything that I'm missing really. So I suggest doing a lot of exercises on this topic and then just moving on. If you can understand the formalism, then you understand it well enough, you just don't realize it.

In my opinion, quotient sets and isomorphism theorems are elucidated a lot by quotient of rings (and of polynomials rings in particular), where you really get the idea that "you make something 0". This is much less clear in groups.

9. Jun 3, 2015

### micromass

As I mentioned, I really recommend you learn the equivalent for rings. It might clear up what we're trying to do! Seeing the constructing of the complex numbers as a quotient makes things very clear.

10. Jun 3, 2015

### Avatrin

Okay, I will try.

11. Jun 3, 2015

### jackmell

Very good. I think that's an important part of the theorem:

Let me ask something then I hope you won't find offensive: you do know why we write $\mathbb{Z}/n\mathbb{Z}$ for $\mathbb{Z}_n$ right? Well, I had a problem with that until I understood the isomorphism theorem: We can construct an onto-homomorphism between the integers and the (additive) group $\mathbb{Z}_n$ simply using the mod function:

$\phi: \mathbb{Z}\to \mathbb{Z}_n, \quad \phi(j)=j\operatorname{mod}n$

then it's clear that the kernel of that homomorphism is the set giving the zero element, or $n\mathbb{Z}$, that is $\phi(n\mathbb{Z})=0$ right?
And since $\operatorname{ker}\phi=n\mathbb{Z}$ then by the isomorphism theorem (and since $\phi$ is onto), we can write $\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}_n$ or since that means they're really just the same group, write $\mathbb{Z}/n\mathbb{Z}$ in place of $\mathbb{Z}_n$.

Edit: had to clarify I'm dealing with additive groups here and not necessary to treat them as rings. Thinks that's correct anyway.

Last edited: Jun 3, 2015
12. Jun 3, 2015

### micromass

And that should give you a hint on how best to define $\mathbb{Z}_0$ or $\mathbb{Z}_{-1}$ (which I guess is not standard). This has as nice consequence that all cyclic groups are of the form $\mathbb{Z}_n,~n\in \mathbb{Z}$, with no exception.

13. Jun 3, 2015

### Avatrin

Well, I did know that, but only due to intuition; It seemed obvious, almost trivial, to me. So, it is interesting to see it in a more rigorous way.

Now, I think I understand the first isomorphism theorem quite well. I only need to understand the two others. The third one seems to be the easiest one to memorize, yet it is the one I find the hardest to understand (cosets of cosets makes every proof look messy).

14. Jun 3, 2015

### mathwonk

the theorem i think you are calling the 3rd theorem (maybe (A/C)/(B/C) ≈ A/B?), just says that you never need to understand cosets of cosets. i.e. if you start from a group and form cosets to creat a quotient group, and then do it again with that quotient, i.e. forming a quotient of a quotient, then you wasted effort and could have obtained the same result just by forming a single quotient of your original group by a bigger siubgroup. Actually this is a trivial corollary of the first isomorphism theorem, since the composition of the two canonical maps from the original group to the second quotient can be consiudered one surjective homomorphism to which you apply the 1st theorem. I.e. the 1st theorem is really the main one. i.e. compose A-->A/C-->(A/C)/(B/C), to get A-->(A/C)/(B/C),

and check the kernel is B. Hence A/B ≈ (A/C)/(B/C). So if you really understand the 1st theorem you get the 3rd one for free. (Also the 2nd one, so the moral here is there is really onloy one isomorphism theorem, the other 2 are "easy" corollaries of it.)

FWIW, here are my free notes with my best shot at explaining this stuff to my 1996 class: see pages 21-26.

http://alpha.math.uga.edu/~roy/843-1.pdf

notice the proof I outlined above of the "3rd theorem" is easier than the one I gave in these notes! We live and learn. Actually if you want to learn this stuff the way I did, then give lectures on it and write up your best attempt at explaining it to someone else, which is what I did in these notes! I.e. I learned them more by writing these notes than by reading anyone else's. I wrote them, then lectured, then came home and rewrote them based on what I learned worked and what didn't in class.

Last edited: Jun 3, 2015
15. Jun 3, 2015

### Avatrin

That is indeed what I call the third isomorphism theorem. I will try to prove it on my own, and if I succeed to some degree, I will post it here for feedback.

16. Jun 3, 2015

### jackmell

I think it's instructive and educationally beneficial to work through a practical example:

$\left(\mathbb{Z}_{13}^*/A\right)/\left(A/B\right)\cong \mathbb{Z}_{13}^*/B$

(I think $A=\{1,3,9\}$ and $B=\{1,3,4,9,10,12\}$ would work)

Say prepare a 15 minute presentation suitable for presenting to your class explaining every minute detail of this problem so that they'll understand it clearly and intuitively enough to do another integer mod group on their own. The act of preparing that presentation will in my opinion more solidify your understanding of the concept.

17. Jun 4, 2015

### micromass

Try to work out what the isomorphisms mean for the group $\mathbb{Z}$. So take the second isomorphism theorem for abelian groups, it states:

$$\frac{S+N}{N} = \frac{S}{S\cap N}$$

So let $S$ and $N$ be subgroups of $\mathbb{Z}$. Do these subgroups have a special form? What do all the subgroups of $\mathbb{Z}$ look like? Under that form, what does $S+N$ and $S\cap N$ mean. To which algebraic identity does the second isomorphism theorem reduce? Do the same for third isomorphism theorem.

Try to do the same for the groups $\mathbb{Z}_n$. See if you can associate the isomorphism theorems to special relations in number theory.

18. Jun 4, 2015

### Avatrin

First, I'll try what I wrote I would do in #15:

I'll start off with mathwonk's suggestion. We have a homomorphism $\phi : G \rightarrow \frac{G/N}{H/N}$ where $\phi(g) = gN \frac{H}{N}$ which is obviously surjective. We know that N and H are normal subgroups of G, and N is a subset of H (why does N not have to be normal in H?). I have to show that:

$$\phi^{-1}(e) = H$$

Well, for $a \in H$ we have that $\phi(a) = aN \frac{H}{N} = \frac{H}{N}$ because aN only permutes the sets in $\frac{H}{N}$. So, I have to show that if $\phi(a) = \frac{H}{N}$, then $a \in H$. Well, if $\phi(a) = aN \frac{H}{N} = \frac{H}{N}$, then $aNb_iN = b_j N$ for all $b_i \in H$. That means that $ab_i = b_j \in H$, and thus, $a = b_j b_i^{-1} \in H$. So, a has to be in H.

Well, I played with it on paper for a while and came up with (for $a\mathbb{Z}, b\mathbb{Z}$):

$$\frac{gcd(a,b)\mathbb{Z}}{a\mathbb{Z}} \simeq \frac{b\mathbb{Z}}{lcm(a,b)\mathbb{Z}}$$

I'll do the third isomorphism theorem later.

Last edited: Jun 4, 2015
19. Jun 4, 2015

### micromass

Right, so by taking cardinalities, we end up with the curious relationship $\text{lcm}(a,b) = \frac{ab}{\text{gcd}(a,b)}$.