Monthly Archives: March 2019

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.)