Proving an Abelian Group with 2 Elements of Order 2 has a Subgroup of Order 4

[188818881888]

Homework Statement

Prove that an abelian group with two elements of order 2 must have a subgroup of order 4

Homework Equations

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.

 

[vela]

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?

 

[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).

 

[188818881888]

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?

Well inorder to show something 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 can't figure out is how to tie that into 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 that's four elements? atleast?

 

[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)?

 

[vela]

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

 

[188818881888]

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)?

that means a= a inverse which is not true

 

[188818881888]

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

ab, a inverse and b inverse

 

[vela]

that means a= a inverse which is not true

Why not?

 

[188818881888]

Why not?

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?