*In which we find that the alternating group is a group, and study subgroups in more detail.*

- 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 the fact from last time 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 . - Theorem 20:
*Let be a cyclic group.**If is finite, with , then is isomorphic to .**If is infinite, then is isomorphic to .*

The first part was clear from our previous work. For the second part, we defined a map that sends to . Checking that this is an isomorphism is an exercise.

### Understanding today’s lecture

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 ?

### Further reading

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.

### 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?

March 6, 2015 at 11:06 am

[…] Expositions of interesting mathematical results « Groups and Group Actions: Lecture 5 […]

May 1, 2015 at 11:15 am

[…] We proved this using the subgroup test. […]