Question

Please help! A4 = {1, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)} is the alternating group, which consists of all even permutations of S4.

Consider the following subset
H1 = {1, (12)(34), (13)(24), (14)(23)}
of S4. Is H1 a subgroup of A4? The same question for the subset
H2 = {1,(123), (132), (124), (142), (134), (143), (234), (243)}

Answer 1

So two questions:
1. Is H1 a subgroup at all?
2. Is it a subgroup of A4?

First H1 is closed under multiplication as every element is of order 1 or 2. Second, every member of H1 is also a member of A4 so yes.

For H2, it is a subgroup, because the elements are all of order 1 or 3, and written pairwise next to their inverse e.g., (123)^2 = (123)^-1 = (132)

Now is it subgroup of A4?

Answer 2

Ok, I'm a little confused. One of the axioms for the definition I have says "for all a in H, a^-1 is in H. So doesn't this mean that H1 is not a subgroup since we can't find all the inverses of all the elements (belonging to H1) in H1?

Answer 3

What's the inverse of (12)(34)?

Answer 4

(21)(43)?

Answer 5

and (21)(43) = (12)(34)

Answer 6

yes?

Answer 7

yes.

Answer 8

So every element in H1 is its own inverse. e of course is of order 1, e^1 = e; all the others are of order 2: e.g., ( (12)(34) )^2 = e

Answer 9

Hence for every h in H1, h^-1 is also in H1.

Answer 10

...because h^-1 = h

Answer 11

Ok. So is H2 also a subgroup of A4?

Answer 12

Go through the axioms
1. Is it a (sub)group: identity, closure under multiplication, existence of inverses
2. Then is it a subset of A4?

Answer 13

Are the answers yes to all those questions?

Answer 14

Yes. i think? Can you explain why you use the order of the cycles to figure out whether it's closed under multiplication?

Answer 15

it's a short-hand really. We don't need to use it, but it's useful to know and after working with these things for enough time, you'll build up more intuition about them and think about them in this among other ways.

Answer 16

ok. Thank you.

Answer 17

Wait, was 'yes' the correct answer?

Answer 18

Yes, H2 is also a subgroup of A4 as you've written it.

Now my question for you: is that really A4? Why is (132) for instance an even permutation?

Answer 19

Honestly, I don't know the answer to that question. The problem was just written that way. I think a permutation is even if the 'sign' is even?

Answer 20

It is A4. Anyway, let's see what your teacher/professor gives you next week. I think s/he is setting you up for something interesting about A4.

Answer 21

ok ... I need to go do my own stuff now.

Good luck with group theory. It's nice stuff; I hope you're enjoying it.

Answer 22

It is pretty interesting. Thanks so much for your help!