Thread: Show that a nonabelian group must have at least five distinct elements.

I have shown that there are distinct elements a and b st I have also shown that there is a unique identity element.

Since G is a group, there a and b must have inverses. I have shown that and What I can not find out is how to show that and Does anyone have any ideas?

Originally Posted by auitto

I have shown that there are distinct elements a and b st I have also shown that there is a unique identity element.

Since G is a group, there a and b must have inverses. I have shown that and What I can not find out is how to show that and Does anyone have any ideas?

I am unsure what you are trying to prove-it seems to me that you are trying to prove that there exists a non-abelian group with 5 elements (which, curiously, there doesn't).

I feel it is probably easiest to attack the problem by looking at the possibilities for groups of order 1, 2, 3 and 4. Try looking at their Cayley tables.

The question is asking me to prove that a non-abelian group has at least five distinct elements. But you're right I guess that I was trying to show that there was one with 5 elements, which is probably why I wasn't getting anywhere. Thanks!

Originally Posted by auitto

I have shown that there are distinct elements a and b st I have also shown that there is a unique identity element.

Since G is a group, there a and b must have inverses. I have shown that and What I can not find out is how to show that and Does anyone have any ideas?

Instead of looking at and , try taking as your five elements, and show that they are all distinct.

Thanks, that worked beautifully!