The Fundamental Theorem of Finite Abelian Groups

It’s been a while since I’ve posted, mainly because I’ve been trying to work through a single theorem: the fundamental theorem of finite abelian groups. I read through the proof as presented in Gallian, and then, because it was so concise and dense, I decided to try to work through the presentation of the proof in Carter’s Visual Group Theory. I had hoped that this second exposition of the proof would be clearer, but in reality, Carter presents the proof as a series of exercises which guide the reader through creating the proof. Unfortunately, while I made it through most of these exercises, the last one was rather opaque, and, although I think I’d have been able to make it through it, I decided that I wanted to move on to the next topic (Sylow theorems) so that I’d have time to spend on Sylow before I have to wrap up the independent study by the end of spring exams.

The statement of the fundamental theorem of finite abelian groups is as follows: given an abelian group GG can be expressed uniquely as the direct product of cyclic groups of prime-power order. In a sense, one can view this theorem as a version of the fundamental theorem of arithmetic (which states that every natural number has a unique decomposition as the product of powers of primes) for abelian groups.

I won’t include every exercise from Visual Group Theory that I worked on since many of them are rather verbose and fairly straightforward. I will include one, however, that is more interesting and that I found more difficult.

This exercise ultimately concludes that, even if the image is not cyclic, we can apply the same homomorphism to the image of the original homomorphism, and continue applying this homomorphism to the image of the previous application of the homomorphism until we finally get the trivial group. The group in this chain of images that comes right before the trivial group must be a cyclic group of order p, and so then part (a) of this exercise then guarantees that all of the groups in the chain are cyclic (by applying the result once for each homomorphism in the chain). This eventually shows that G itself must be cyclic. Thus, part (a) is the heart of the exercise—the rest of the parts just deal with defining the chain of homomorphisms and groups that eventually allows us to apply the result of part (a).

The overall chain of reasoning of the proof is as follows: First, we show that any abelian group can be decomposed into the direct product of (not necessarily cyclic) abelian groups whose orders are powers of primes. We then show that any abelian group whose order is a power of a prime into the direct product of cyclic groups of prime-power order. The exercise I described above establishes a crucial fact that is instrumental in the second part of the proof.


Homomorphisms and the Isomorphism Theorems

In our most recent meeting, we discussed a pretty hefty chapter that covers homomorphisms and the first isomorphism theorem (along with some other results), which is one of the more important theorems from group theory. Then we worked on two exercises from the chapter that asked us to use the first isomorphism theorem to prove the second and third isomorphism theorems.

To begin, I’ll just include a statement of the first isomorphism theorem:

Now, we were tasked with the following two exercises:

Our line of reasoning for the second isomorphism theorem was as follows:

And for the third:


Cosets and Lagrange’s Theorem

I’m just now getting around to writing this post, although I worked through these exercises last week. These exercises come from the chapter on cosets and Lagrange’s theorem.

My solution is as follows (included as an image so that I could more easily include well-formatted math):

The exercise that we couldn’t solve is this one:

Although it looks fairly straightforward, it really isn’t, at least using any methods that I know so far. We struggled to figure out what facts or theorems to take advantage of in order to prove the desired result. (I even tried looking up solutions online, and all of them rely on results that I haven’t proven yet.) This highlights how often times the most difficult part of solving an exercise isn’t executing a proof once you have a plan of attack, but coming up with the plan of attack in the first place.


Isomorphisms and Cosets

Just an hour ago, I had my most recent meeting with Dr. Prudhom, in which we discussed several exercises from the chapter on isomorphisms and several more from the chapter on cosets and Lagrange’s theorem. Most of them were fairly straightforward, except for several that were a bit trickier—there was even one that we still haven’t figured out.

I’ll begin with the ones from the isomorphism chapter, which were overall easier than the coset exercises. The first group of exercises that I tackled formed a little sequence, where each one built on the previous ones:

58 was not particularly difficult. A solution very similar to the one that I found can be found in the answer to this Math Stack Exchange question. 59 was tricker since it required that I find a specific example of a proper subgroup of the rationals that is isomorphic to the positive rationals under multiplication. I ended up showing that the set of all rational numbers of the form m/n, where both m and n are perfect squares (and positive), with the operation of multiplication is a subgroup of the positive rationals—I called this subgroup G. Then, I showed that the mapping that sends the rational number m/n to the element m^2/n^2 of G is an isomorphism. Here’s my full solution:

Exercise 60 is simple once we solve exercise 58. Although exercise 58 specifically discussed automorphisms of the rationals, the proof that I provided for it actually works for any arbitrary homomorphism of the rationals, so it also applies to isomorphisms of the rationals onto a proper subgroup of itself. My solution involved supposing that there was a proper subgroup of the rationals under addition that was isomorphic to the rationals under addition, and then showing that such a subgroup actually had to contain every single rational number, and thus was not a proper subgroup.

The exercise that I had the most trouble with was 47:

(The inner automorphism induced by x maps each of element a of the group to the element xax^-1.)

Here is my complete solution:

I was able to deduce fairly quickly that x and y must commute, but I had trouble going from that point to figuring out what this said about x and y as permutations.

This post is getting rather long, so I’ll leave the coset exercises that I did for the next post. (That’s the chapter that had the exercise that we still haven’t figured out how to solve.)

An Interesting Connection: Fundamental Groups

While I’m still continuing to work through Gallian, I recently stumbled across an interesting mention of group theory in the textbook (The Shape of Space, by Jeffrey Weeks) for my geometry and topology class this year. It mentioned that a number of techniques and tools can be used to determine if surfaces are in fact the same surface or not. This is not always immediately apparent, because surfaces can have different representations that can look completely different upon inspection. For example, the hexagonal 3-torus and the regular 3-torus are in fact topologically the same, even though they appear to be very different from their definitions.

One of the tools that can be used is the fundamental group, which can provide information about the properties of a surface. While there’s certainly no way that I could understand the formal definition of the fundamental group right now, given that it requires knowledge of formal definitions in topology (which our textbook does not include—instead, we’ve been taking an intuitive approach to the subject), I was able to find an intuitive definition of the fundamental group of a topological space on Wikipedia that actually does make sense to me. This intuitive definition reads:

Start with a space (e.g. a surface), and some point in it, and all the loops both starting and ending at this point — paths that start at this point, wander around and eventually return to the starting point. Two loops can be combined together in an obvious way: travel along the first loop, then along the second. Two loops are considered equivalent if one can be deformed into the other without breaking. The set of all such loops with this method of combining and this equivalence between them is the fundamental group for that particular space.


In short, the fundamental group of a space is the set of the equivalence classes of closed loops (starting and ending at some given point) in the space under the equivalence relation where two loops are equivalent if one can be continuously deformed into the other. The operation of the group is basically the concatenation of the two paths: trace one path and then trace the other. The resulting path is still a closed loop because it starts and ends at the same point and is continuous because the first two loops were both continuous and closed. Thus, the fundamental group is indeed a group.

Some examples: the fundamental group of the sphere is the trivial group (the group containing only the identity and nothing else) because any closed path can be continuously deformed into any other closed path. The fundamental group of the cylinder is isomorphic to the group of the integers under addition, where each equivalence class of loops is mapped to an integer based on how many times it wraps around the cylinder. If a loop wraps around twice, then it can’t be continuously deformed into a loop that wraps around only once. But any two loops that wrap around the same number of times are equivalent under this equivalence relation.

It can be sort of tricky to figure out a space’s fundamental group intuitively—the Klein bottle is one that pushes the limits of this intuitive definition—but I thought that it was cool that group theory was referenced in a book on intuitive topology, highlighting how group theory is at once both a field of study in and of itself as well as a tool that is leveraged in many other areas of mathematics.

Permutations and Permutation Groups

Yesterday I had a meeting with Dr. Prudhom in which we discussed a few exercises from the chapter on permutations and permutation groups—the next important type of group after the dihedral groups and cyclic groups. One key topic that is necessary to understand permutation groups is cycle notation for permutations, which I’ve seen before but always struggled with. This time around, I spent a fair amount of time working specifically on cycle notation just to be sure that I understood it, because it’s the notation that is used whenever permutations are involved.

The exercises that we discussed in the meeting were the following:

For 50, we ended up trying out some examples and realized that whenever you conjugate a permutation by a 2-cycle, there are two possibilities. Either the two permutations are disjoint, in which case they commute and so you get that the 2-cycle just cancels with itself. If, however, the permutations are not disjoint, then we worked through the example as follows.

We can write the situation like (ab)(….a…b….)(ab), where the middle permutation is a t-cycle involving a and b at some point. When we work through this, we see that the a gets mapped to the element right after b in the t-cycle and b gets mapped to the element right after a in the t-cycle. Furthermore, the element right before a in the t-cycle gets mapped to b, and the element right before b in the t-cycle gets mapped to a. So this means that a and b have their positions swapped in the t-cycle when we conjugate by a 2-cycle, that is, (ab)(….a…b….)(ab)=(….b…a….). This is plainly still a t-cycle, so we’re done. (There’s technically also the case where a and b are adjacent in the t-cycle, but this doesn’t change anything. And in the case where only a or only b appears in the t-cycle, but not both, then the result is just replacing a with b in the t-cycle and holding the other one fixed. In both scenarios, the result is still just a t-cycle.)

51 follows fairly quickly from 50: we can assume that the cycles that form b are all disjoint, so they commute. Then we can write a as a product of (not necessarily disjoint, in this case) 2-cycles. These 2-cycles will commute with any cycles of b with which they are disjoint, so we can move them to be adjacent to the cycles of b with which they share elements. Then we can apply the previous exercise multiple times in order to collapse all of the 2-cycles into the cycle of b with which they share an element. In the case that there’s a 2-cycle from a that shares one element with one cycle of b and one element with another cycle of b, then we have something that looks like (ab)(….a…..)(….b….)(ab). This sends b to the element after a, a to the element right after b, the element right before a to the b and the one right before b to a. In other words, (ab)(….a….)(….b…..)(ab)=(….b….)(….a….). This doesn’t change the length of either of the cycles. Thus we can collapse all of the 2-cycles of a into the cycles of b without changing the length of the cycles of b, which is sufficient to complete the exercise.

And 52 follows from 51: we just showed that conjugating a permutation by another permutation doesn’t change the length of the disjoint cycles that make up the inner permutation. Therefore, conjugation doesn’t change the parity of the inner permutation either. (I also found an alternate proof that uses a slightly more abstract approach here—see Corollary 2.11 and the preceding theorem. I like this approach as a cool “other method” to approaching the problem.)

Cyclic Groups

I’m quite busy this week, and so I’ve basically accepted that I am going to fall behind the plan that I laid out in my proposal. I hope to be able to work more quickly in the coming weeks to make up.

I had a meeting on Monday, in which we discussed several exercises from the chapter on cyclic groups. Most of the exercises from this chapter were fairly straightforward once you figure out how to approach them. In other words, they require very few steps to solve, but finding the correct angle of attack can be a challenge. For example, an exercise that we had some trouble with is as follows:

The key here is to observe that the order of every element of a cyclic group divides the order of the group, so if we raise an element to the p^n – 1 power, we should get the identity of the group. Then, if we raise an element to the p^n power, we should get the original element back. If we call the group in question G, then, notationally,

Note that p^n=p*p^(n-1), so we can write

So we have written a as a pth power of an element of the group. Because a is just an arbitrary element of the group, it applies to every element of the group, and so we are done.

This exercise and its solution highlights how the solutions often take advantage of basic divisibility facts, which is why they tend to be fairly straightforward once you find the proper method, but difficult until you manage to find the proper method.



In the last week of January and the first several days of this week, I worked through the chapter on subgroups and began the chapter on cyclic groups. I found most of the exercises from the subgroups chapter to be fairly straightforward, except for a couple that we discussed in our last meeting. I’ve included the interesting one below.

Finding the orders of the elements (in this notation, |a| denotes the order of element a) was not a challenge—in fact, it’s really a mechanical process. The tricky part was finding the relationship between the orders, like the last part of the question prompts. The orders are as follows:

  • |6|=2, |2|=6, |6+2|=3
  • |3|=4, |8|=3, |3+8|=12
  • |5|=12, |4|=3, |5+4|=4

There are some patterns that stand out right away, but none are quite precise enough to be the relationship that the book is likely looking for. For example, note that in every set, the orders of two of the elements multiply to give the order of the third. The issue is that there is no pattern to which ones multiply together and which one is given by multiplication of the other two. Another pattern is that all of the orders divide the order of the group. But this isn’t a relationship between the orders of ab, and a+b, and is also fairly straightforward, and is covered in the next chapter on cyclic groups (which I’m working through right now).

We also looked for a relationship involving some sort of greatest common divisor involving 12, a, and b, but couldn’t find anything there. We ended the meeting thinking that the pattern perhaps was that |a+b| divides |a||b|in other words, that the order of the sum divides the product of the orders.

I did some more research after the meeting on my own, looking into whether or not, given the orders of a and b, we can find the order of ab (note here I just switched to multiplicative notation, but it’s really the same as a+b was in the integers mod 12, because I’m just omitting the group operation). It turns out that we can’t, in general. we can’t say anything about the order of the products. I can’t understand the proof, because it relies on ring theory and some basic linear algebra (involving groups of matrices), but maybe by the end of the study I will be able to tackle it… Regardless, the result I found is Theorem 1.64 in this document, which I encountered via this MSE post.


A Solution to Exercise 42

I mentioned in my last post that we had discussed exercise 42 from chapter 2 in our last meeting and that Dr. Prudhom gave me a hint on a tactic I could take to solve it. I tried using that method, but I found it to be actually much more difficult than I had anticipated—maybe I went about it in the wrong way, but the algebra was incredibly messy and felt much more convoluted than necessary. So I came back to the original statement of the exercise and tried to approach it from a group-theoretic perspective.

The first thing I did was to look up some examples of dihedral groups online, using a very handy website I found called GroupProps, which is basically “Wikipedia for groups.” You can find pages on a huge variety of groups that describe their properties and definitions. I looked up the dihedral groups, and specifically, the dihedral groups with orders that are multiples of 4 (i.e. the symmetry groups of regular polygons with an even number of vertices). I arrived at this restriction because I noted that the exercise implies that a rotation by 180° is an element of the group, and this is only the case in these specific dihedral groups. After all, if you rotate a triangle by 180° about its center, you don’t end up with a symmetry.

I noticed that in their presentations of these groups, GroupProps lists each reflection as a product of some “base” reflection with some number of rotations. That is, there are two basic, essential elements to the group, from which every other element can be constructed. (These are called a generating set of the group.) This led me to believe, although I have not been able to prove it, that any reflection in the dihedral group of order 2can be written as the product of a single reflection x and some number of rotations by 360°/k degrees. (Call one rotation of 360/k° a.)

This observation lets me restate the problem (working in the dihedral group of order 2k) as follows:

I can then manipulate this expression into this form without changing the value of either side of the equation:

Note that I multiplied both sides by the inverse of x^2. But x^2 is the identity, so its inverse is also the identity, so I actually didn’t change the value of either side of the equation! This is important. (Just as a note, this works specifically because x is its own inverse, so I could have just replaced x with x^-1 without further justification. But this makes it even clearer that this step is justified.)

Once I have it in this form, I can use a second observation I gleaned from the GroupProps website, specifically, that

(Again, I don’t know how to prove it, but it makes intuitive sense if you play around with some concrete examples to see it in action.)

Edit (Jan. 31): I did find a way to prove it, and it’s actually really easy and I don’t quite see how I missed it. There’s absolutely no trick to it, all you have to do is write out the multiplication:

The last equality holds because the xs cancel with the x^-1 s that are right next to them.

That fact lets me rewrite the equation as follows, again, without changing the value of either side:

Now I just have to solve the modular congruence n-m=m-n mod 2k, which I’m comfortable doing thanks to my GOA Number Theory course last semester. I get that either n=m mod 2k, or n-m=m-n=k mod 2k. But the first possibility is actually not a possibility, because we assumed that the two rotations were distinct, and this would imply that they are the same rotation. So we have that n-m=m-n=k mod 2k.

Recall that I never changed the value of either side of the equation through all of my manipulations. Considering just the left hand side of all of the equations, this means that I have

My notation was different from the book’s notation, but this equality is exactly what I was asked to prove. I included the book’s notation under my equation for clarity.

I’m not sure that I took the most direct method, and I know that there are a lot of “holes” in my reasoning, insofar as there are facts that I used that I haven’t seen proven. But I think that this method does work regardless.


Introduction to Groups (Chapters 1-2) and First Meeting

I had my first meeting for this independent study yesterday (Friday, Jan. 25), in which we discussed several exercises from chapter 2 of Gallian. (Chapters 0 and 1 focused primarily on fundamental prerequisites, like basic number theory and techniques of proof, as well as on an intuitive exposition of groups.) The exercises that we discussed assumed only basic knowledge of the definition of a group, some essential properties of groups, and understanding of one specific type of group (dihedral groups). We fully solved two exercises, and Dr. Prudhom gave me a hint on a third that I’m now trying to solve on my own.

The two exercises that we ended up solving entirely were numbers 17 and 39, which read as follows:

While I did learn the solutions to the exercises, which improved my understanding of groups, I found the process of solving the exercises to be most interesting. For 17, I had expected there to be some “trick” to solving the problem that made it quick and easy once you saw it, but it turned out that it just required some careful thought. In other words, we had to be willing to experiment and try things out in order to eventually solve the exercise. The same was true for 39; we spent a while trying various approaches to the problem until we finally encountered the correct one by looking online for hints. If I remember correctly, the hint that we found suggested finding some x such that axb=bxa in any, possibly non-Abelian group, from which it would follow from the premise of the exercise (specifically, that axb=cxd implies ab=cd) that ab=ba. Once we got the hint, it was just a matter of making educated guesses at what a suitable x would look like.

Finally, we also discussed a third exercise, which I’m now trying to solve using the hint that Dr. Prudhom gave me. This exercise was as follows:

The nth dihedral group is the group of symmetries of a regular n-gon. Before the meeting, I had made an educated guess that the only reflections that commute with each other are ones with perpendicular axes of reflection. If this turns out to be the case, then it follows that the composition of the two reflections is the same as a rotation by 180°. Dr. Prudhom suggested proving this not using group theory, but rather with coordinate plane geometry. That is, he suggested that I try to show that if the result of reflecting a point over the line y=mx and then over the line y=nx is the same as reflecting first over the line y=nx and then the line y=mx, then n=-1/m (which means that the lines are perpendicular). This general statement implies the group-theoretic version because we can think of the vertices of a regular n-gon as points on the coordinate plane.

Again, the approach that he suggested to this problem was quite interesting, and highlights how problems can be “translated” into an equivalent form that’s easier to solve. In this case, as opposed to solving the original group-theoretic problem, we translated it into a problem using basic algebra and geometry, topic with which I’m much more comfortable.