any mathematician out there? i need help?

Question

i worked on this for over 2 hours now, plz help
suppose G is a group that has exactly 8 elements of order 3. how many subgroups of order 3 does G have?

all info are in the question
its not insufficient data

Answer 1

There are 8 elements of order 3 and each subgroup of order 3 has 3 elements (2 of order 3 and 1 of order 1). Each subgroup must be closed under inverses and each of the elements have a unique inverse. Also since each subgroup is of order 3 (a prime) they have a trivial intersection. So this means that there are 8/2=4 subgroups of order 3.

By the way, math problems quit often take longer than 2 hours...I have been working on some problems for weeks and I haven't stopped yet.

Note on pevious person's answer: Two elements of order 3 multiplied together do not necessarily give an element of order 3, if the group is nonabelian, we have no idea what the order is without more information. Also if the elements of order 3 multiply to give an element of order 3, then that element would be one of the 8 mentioned, so it is already accounted for. Also you should note that two elements of order 3 could multipy to give an element of order 1 (the element and its inverse).

Answer 2

I think it's 5. You need the identity in any subgroup. Take an element of order three to be in the subgroup. Call it A. A^2 must be another element of the subgroup, because it is distinct from A and e (identity). A^2*A=e. A^2*A^2=A^2*A*A=e*A=A. So these are all the elements you need for a subgroup.

Also note that A^2 is of order 3. So the elements of order three can be paired off. There are four pairings, so there are 4 subgroups here.

I don't think there's a guarantee if you pick two elements of order three, that you'll get a subgroup.

However, all of the elements of order three will be a subgroup. If you multiply any two of them together, they are of order three, and they are in the group, so they must be in the subgroup of all elements of order 3.

That is, A^3=e, B^3=e, (AB)^3=A^3B^3=ee=e. Since AB is of order three, it is some element of G with order three, so you already have it in your subgroup.

Hope that helps.

Answer 3

24

Answer 4

I think you might be able to have 24 subgroups.

Answer 5

This link says there are 4. Too tired to follow it completely though.

Answer 6

insufficient data