Math Is Fun Forum

deepz

Member

Registered: 2007-03-20

Posts: 10

Algebra help needed

Hi i need help with this question:

Let G be the symmetric group S3.
a)What are the orders of elements of G?
b)Show that G contains just one subgroup of order 3 and three subgroups of order 2

Thanks in advance

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Algebra help needed

S[sub]3[/sub] has six elements: 1, (12), (13), (23), (123), (132).

Their orders are 1, 2, 2, 2, 3, 3 respectively.

The subgroup of order 3 contains 1, (123), (132). The subgroups of order 2 are {1,(12)}, {1,(13)}, {1,(23)}.

deepz

Member

Registered: 2007-03-20

Posts: 10

Re: Algebra help needed

thanks alot!

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Algebra help needed

Their orders are 1, 2, 2, 2, 3, 3 respectively.

In general, a k-cycle has order k.  So (12) is a 2-cycle, and thus, has order 2.

The subgroup of order 3 contains 1, (123), (132). The subgroups of order 2 are {1,(12)}, {1,(13)}, {1,(23)}.

I'm not sure that fully answers the question.  Mustn't we also show it isn't possible to have any more subgroups of order 3 and 2?

For any subgroup of order 3, as it is prime, it must be cyclic and it can be generated by any of it's elements.  Thus, each element in a subgroup of order 3 must have order 3 (other than the identity).  The one subgroup of order 3 is {1, (123), (132)}.  If there is another subgroup of order three, then there must be some element of S3 that is not in here and of order three.  The previous question shows that such an element does not exist, and so there are no more subgroups of order three.

This same logic can be applied to subgroups of order 2, and in fact, and subgroup of order p.

But perhaps this is overkill.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

freddogtgj

Member

Registered: 2006-12-02

Posts: 54

Re: Algebra help needed

Wouldn't the elements be:
(123)          (123)          (123)
(123)          (132)          (213)         

(123)          (123)          (123)
(231)          (312)          (321)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Algebra help needed

I'm not sure what you're trying to say with the grouping, freddogtgj.

But notice that (123), (231), and (312) are all the same exact element, as p(1) = 2, p(2) = 3, and p(3) = 1 for each of them, so they are all the same permutation (sends the inputs to the same places).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."