Counterexample involving an abelian group

[calvino]

Basically, I have to show an example such that for a nonabelian group G, with a,b elements of G, (a has order n, and b has order m), it is not necessarily the case that (ab)^mn= e. where e is the identity element.

im not sure where to start. =\

 

[AKG]

What kinds of groups do you know about? What have you tried so far? I found that choosing m and n appropriately, I was able to think of a lot of examples very quickly. So maybe try thinking about what m and n you should choose.

 

[calvino]

well... we've dealt with a lot of permutations- rotations/reflections on fixed objects, planes about a point, cycles... but i suppose since we know the definition of a group that it shouldn't matter what group(s) we choose to use in this question. I've basically tried cycles - like

a=(123), b=(12)...and well..it works out. My problem lies in the fact that I cannot computer compositions fast, and so I'm discouraged when I have to calculate any exponents. You said that you thought of some rather quickly, yet to me it's very difficult to just "think" them up. Perhaps there is something I'm missing, because you said for me to just pick appropriate m and n, and I have no clue how to do so, since they relate to specific permutations. I guess I'll just study for a bit, until I know my stuff enough to post again. Thanks for your help.

 

[calvino]

i think this works:

(12), and (13), in S3.

(12) has order 2, and (13) also has order 2, yet (13)(12)=(123), and (123)^4 does not equal (1)(2)(3). Where I went wrong was with the fact that i forgot that all cycles abelian (at least that's where i believe i went wrong).

Now, I noticed that (123) squared is (321), and cubed is (123) again. Is this so for all cycles? Furthermore, are there any other sort of "tricks" to know when generating groups, and doing powers of elements of groups?

 

[AKG]

Well I found it was very easy when I let m = n = 2. You've dealt with rotations/reflections, and that's the first thing that came to mind. Dealing with 2-dimensional space, two reflections (each reflection has order 2) composed with one another gives a rotation (so long as the lines are not parallel; if the lines coincide, the composition of the two reflections gives the trivial rotation - identity). If you pick the angles of your lines right, you can make a rotation of any order, including infinite. From this example, I was able to quickly think about the dihedral groups (rotational and reflectional symmetries of the n-gons) and again just look at the composition of two reflections. Of course, you can then find an example from the wall-paper groups, etc. Now one thing (I believe it's Cayley's theorem) states that every finite group is isomorphic to the subgroup of a permutation group, specifically, if G is a finite group of order n, then G is isomorphic to a subgroup of S_n. So sticking with the theme of elements of order 2, I took two transpositions as you did, and knowing that I could create a 3-cycle from them if I chose them right, as you did, I would get something x such that x⁴ is not identity.

There are a lot of good tricks when it comes to permutation groups. First, the permutation group S_n is generated by the transpositions (2-cycles). Also, you should know that an arbitrary permutation can be decomposed into the product of disjoint cycles. Also, the cycle (a₁ ... a_k) = (a₁ a_k)(a₁ a_k-1)...(a₁ a₂). The alternating subgroup A_n (the even permutations, i.e. those whose decomposition into a product of transpositions always has an even number of transpositions, no matter how you decompose it) is generated by the 3-cycles. A k-cycle has order k.

Anyways, I picked m=n=2 because that's the smallest possible number, and I figured it would be easiest to work with.

Where I went wrong was with the fact that i forgot that all cycles abelian

What's that supposed to mean? It's a typo I guess.

Now, I noticed that (123) squared is (321), and cubed is (123) again.

This is not true. (123)² = (321). (123)³ = e (identity). (123)⁴ = (123)³(123) = e(123) = (123) which is not e. Note that this agrees with what I previously mentioned that a k-cycle has order k, since we see the 3-cycle (123) is such that (123)³ = e.

A lot of the tips and tricks I mentioned about permutation groups should be in any book that covers the permutation groups. Such a book should also have a few problems asking you to compute the product of some permutations. I suggest you do that, so that computing them becomes easy. e.g.:

(143)(32)(87)(567)(1458)

Start by looking at 1, and go from right to left. So first, 1 goes to 4. Next (looking at (567)) 4 doesn't move. Next (looking at (87)) 4 doesn't move. And again, 4 doesn't move. Finally, 4 goes to 3, so your answer will start out looking like:

(13...

So start with 3, and again go from right to left. 3 doesn't move for the first 3 cycles, then 3 goes to 2, then 2 stays fixed in the last one, so:

(132...

2 -> 2 -> 2 -> 2 -> 3 -> 1, so we get:

(132)(...

Start with the next smallest number, 4:

4 -> 5 -> 6 -> 6 -> 6 -> 6, so we get:

(132)(46...

6 -> 6 -> 7 -> 8 -> 8 -> 8

(132)(468...

8 -> 1 -> 1 -> 1 -> 1 -> 4

(132)(468)(...

The next smallest number is 5:

5 -> 8 -> 8 -> 7 -> 7 -> 7

(132)(468)(57...

(now you should be able to tell that 7 ends up going to 5 without computing anything, just because you can tell the thing has to end now, but just in case):

7 -> 7 -> 5 -> 5 -> 5 -> 5

(132)(468)(57)

If your book doesn't have problems, make some up for yourself, that's what I just did here.