1. ## 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?

2. 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).

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