In which we find that the alternating group is a group, and study subgroups in more detail.
- Theorem 15: (i) Any permutation in can be written as a product of transpositions. (ii) A permutation cannot be both even and odd. We proved (i) by noting that since any permutation can be written as a product of disjoint cycles (Theorem 9), it suffices to show that any cycle can be written as a product of transpositions. And we have . For (ii), we showed that if can be written as a product of transpositions, then .
- Definition of the alternating group .
- Proposition 16:
- is a subgroup of .
- For , the order of is .
- For , is non-Abelian.
Proving this was fairly straightforward. To check the first part, we used our observation that is even if and only if . For the second part, we wrote down a bijection between and . And for the third part, we noted that and are even permutations that don’t commute.
- Proposition 17 (Subgroup test): Let be a group. The subset is a subgroup of if and only if is non-empty and for all , . We noted that if is a subgroup then since it contains the identity it is non-empty, and since it contains inverses and is closed under the group operation we must have whenever , . For the reverse direction, we said that if then so contains the identity, and then for any we have so contains inverses. Then for , , we have and so , so is closed under the group operation.
- Proposition 18: Let be a group. Let , be subgroups of . Then is a subgroup of . This is an exercise on Sheet 3.
- Definition of the subgroup generated by a subset of a group . Definition of the elements of as the generators of .
- Division algorithm: Let , be integers with . Then there are unique integers and such that and .
- Proposition 19: Let be a group. Take .
- We have .
- If has finite order, then .
In each part, it was clear that contains the other set. For the first part, we used the subgroup test to show that is a subgroup of , and then since is the smallest subgroup containing , we get equality. For the second part, we used the division algorithm to show that any (with ) is in .
Understanding today’s lecture
Theorem 15 doesn’t claim that the number of transpositions is well defined, just the parity of the number of transpositions. If I can write a permutation as a product of 6 transpositions, then I may well be able to write it as a product of 8 transpositions (in fact, I can: write it as those six transpositions followed by ). But I’ll never be able to write it as a product of 7 transpositions, or 101.
Can you show that and are not conjugate in ? Can you list the elements of ? (Well, don’t list them, there are lots, but list the cycle types and say how many elements there are with each cycle type.)
Can you prove Proposition 18 (the intersection of two subgroups is a subgroup)? Can you extend it to arbitrary intersections of subgroups (as mentioned in the remark following Proposition 18 in the lecture)?
In , what is the subgroup ? What is the subgroup ?
You might like to find a book and look up the classification of conjugacy classes in the alternating group: it turns out to be rather interesting. We’ll see more about this later in the course.
On a completely separate note, here is a new Polymath project that seeks to prove a problem in linear algebra. This is accessible to first-year undergraduates, so you might like to take a look, you can watch mathematicians tackling an unsolved research problem, and potentially even join in yourself!
Preparation for Lecture 6
If is a cyclic group and is a subgroup of , must itself be cyclic?
We saw that is not cyclic. Is ever cyclic?
We’re going to need to talk about equivalence relations soon, so this would be a good time to remind yourself what an equivalence relation is. Do you have some favourite examples of equivalence relations?