# Abelian

• Oct 21st 2008, 11:11 AM
bluejay
Abelian
Prove that a group of order 4 is Abelian.

To prove that a group is Abelian, I need to show that it has an identity, inverses, is associative, and is closed under the operation. I can definitely do this for specific groups, but I am not sure how to do this for a generality. Can someone please help me get started?
• Oct 21st 2008, 11:40 AM
Tony2710
Couple of thoughts:

all groups of order 4 are either cyclic groups or klein groups.

If an element of the group has order 4 then the group is cyclic, or if all the elements are self-inverse then the group is klein.

This might help you start thinking about proving that all groups of order 4 are abelian (where g*h=h*g).
• Oct 22nd 2008, 05:10 PM
ThePerfectHacker
Quote:

Originally Posted by bluejay
Prove that a group of order 4 is Abelian.

To prove that a group is Abelian, I need to show that it has an identity, inverses, is associative, and is closed under the operation. I can definitely do this for specific groups, but I am not sure how to do this for a generality. Can someone please help me get started?

Let the elements be $\{e,a,b,c\}$. Now the orders of $a,b,c$ must be larger than $1$ and divide $4$ - the order of the group. If any of them are $4$ then the group is cyclic and therefore abelian. Thus, it is safe to assume that $a,b,c$ all have order $2$. Therefore, $a^2=b^2=c^2 = e$. Consider $ab$. It cannot be the case that $ab=b,ab=a$ for that would imply $a=e$ or $b=e$. It also cannot be the case that $ab=e$ for that would mean $b$ is inverse of $a$ - however this is impossible since $a$ is its own inverse and inverses are unique. Thus, it collows that $ab=c$. Using a similar argument we can show $ba=c$. And by symmetry of these elements we have $ac=b,ca=b,bc=a,cb=a$. This shows the group is abelian.