# Homework Help: Abelian Groups

1. Sep 13, 2010

### 188818881888

1. The problem statement, all variables and given/known data
Prove that an abelian group with two elements of order 2 must have a subgroup of order 4

2. Relevant equations

3. The attempt at a solution
Let G be an abelian group ==> for every a,b that belong to G ab=ba.
Let a,b have order 2 ==> a^2 =e and b^2 = e. Since a belongs to G aa=a^2 belongs to G. Since b belongs to G bb= b^2 belongs to G. IE four elements ie order of a subgroup can be four.

2. Sep 13, 2010

### vela

Staff Emeritus
It's not enough to say there are four elements in G to prove that there is a subgroup of order 4. (Plus since a2=b2=e, you've only shown there are three elements in G.) Think about what "subgroup of order 4" means. What do you need to show to say that a subset of G is a subgroup of G and that its order is 4?

3. Sep 13, 2010

### snipez90

Um, what four elements have you singled out? I don't see how saying a^2 belongs to G and b^2 belongs to G helps, since we already know that both are equal to e, which obviously belongs to G since G is a group (and that's a single element).

4. Sep 13, 2010

### 188818881888

Well inorder to show someting is a subgroup you have to show that it is closed under the operation and that if a belongs to the subgroup then a^-1 (inverse of a) belongs to the subgroup. but what i cant figure out is how to tie that in to proving the sungroup has four elements in it. well wait. if a belongs to the subgroup, a inverse should belong there as well. the same goes for b and b inverse. so i guess thats four elements????? atleast???

5. Sep 13, 2010

### snipez90

Again, you have to be careful here. If a^2 = e, what does this tell you about a^-1 (Hint: multiply both sides of the equation by a^-1)?

6. Sep 13, 2010

### vela

Staff Emeritus
Suppose H is a subgroup that includes a and b. What other elements have to be in H?

7. Sep 13, 2010

### 188818881888

that means a= a inverse which is not true

8. Sep 13, 2010

### 188818881888

ab, a inverse and b inverse

9. Sep 13, 2010

### vela

Staff Emeritus
Why not?

10. Sep 13, 2010

### 188818881888

because in a group a(a^-1) must equal e. but if a^-1=a then a^2=e. ok so a should equal a^-1????