Even permutaions and Alternating group help!!

• Apr 8th 2010, 09:01 AM
killercatfish
Even permutaions and Alternating group help!!
I fear I have become stumped with permutations once we hit cycles. I understand cycle notation, and multiplication. I desperately seek help on a few homework problems that I just cannot find footing on!

1. Show that A8 contains an element of order 15.
I thought that since n=8 the largest element could be 8?

2. Prove that (1234) is not the product of 3-cycles.

3. In S4, find a cyclic subgroup of order 4 and a noncyclic subgroup of order 4.

I really do not know where to start on these, the book is quite unhelpful.

Thank you!
• Apr 8th 2010, 09:34 AM
qmech
A8 is a fairly large group. It works on 8 objects (1,2,3,4,5,6,7,8), but it has more elements (permutations). I think it has 8!/2 elements, so the largest order is much larger than 8.

1. What is the order of (123)? Apply this 3 times and you get the identity, so the order is 3. So what is the order of the element (45678)? And what is the order of (123)(45678)?

2. The order of (1234) is 4 and 4 is not divisible by 3. (Verify this suggestion with somebody else).

3. In S4 you have 4 objects (1,2,3,4) to work with. What is the order of the element (1234)? How about the group formed by the 2 elements (12) and (34)? Is the latter group formed by either element by itself? If not, is it cyclic?
• Apr 8th 2010, 10:25 AM
killercatfish
Awesome!

Am I correct in my assumption from your post that the order of an element in a permutation is the number of elements in the cyclic notation? thus (1 2 3 4) has order 4? My book does not have this in there anywhere!

for number 3, I need to find the correct permutation of (1 2 3 4) which gives me the entire set of S4, as well as one that does not, correct?
• Apr 8th 2010, 10:58 AM
qmech
The number of elements in the cycle (123) is 3 and the order is 3.
Yes, for cycles (123...n) the number of elements is n and the order is n, too.
Although this rule works, it is better for you to think about what is happening in the cycle.

For the group element g=(123), the element 1 goes to 2, the element 2 goes to 3, and the element 3 goes to 1. If we just look at element 1, g(1)=2. Then g(g(1))=3 and g(g(g(1)))=1 and it is the number of times we apply g to get back to where we started that defines the order of the element g.

For number 3 you are not trying to get the entire set S4. Rather think of the element h in S4 defined by h=(1234) and ask yourself h(h(h...h(1))))=1; just how many times do you have to use h to permute the element 1 to get 1 again? Remember an element of S4 is not the object 1 or 2 or 3 or 4, rather an element of S4 is a permutation of the objects 1,2,3,4.
• Apr 8th 2010, 12:21 PM
killercatfish
Thank you for all this clarification!

so (1 2 3 4) is an element of order 4 of S4.

That means that (1 3)(4 2) would also be an element of order 4 in S4.
• Apr 8th 2010, 02:28 PM
qmech
Actually g=(1 3)(4 2) is an element of order 2. Maybe this way of writing will be clearer to you:

g acts on the first row to give the second row.
1 2 3 4
3 4 1 2

If you do g again, you'll find everything is back in their original place.

The actual subgroup has 4 elements:
the identity
(13)
(24)
(13)(24)

The identity has order 1, and each of the others have order 2.