# Math Help - Show that a nonabelian group must have at least five distinct elements.

1. ## Show that a nonabelian group must have at least five distinct elements.

I have shown that there are distinct elements a and b st $ab\neq ba.$ 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 $a^{-1}\neq e, b^{-1}\neq e, a^{-1} \neq b$ and $b^{-1}\neq a.$ What I can not find out is how to show that $a^{-1}\neq a$ and $b^{-1}\neq b.$ Does anyone have any ideas?

2. Originally Posted by auitto
I have shown that there are distinct elements a and b st $ab\neq ba.$ 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 $a^{-1}\neq e, b^{-1}\neq e, a^{-1} \neq b$ and $b^{-1}\neq a.$ What I can not find out is how to show that $a^{-1}\neq a$ and $b^{-1}\neq b.$ 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.

3. 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!

4. Originally Posted by auitto
I have shown that there are distinct elements a and b st $ab\neq ba.$ 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 $a^{-1}\neq e, b^{-1}\neq e, a^{-1} \neq b$ and $b^{-1}\neq a.$ What I can not find out is how to show that $a^{-1}\neq a$ and $b^{-1}\neq b.$ Does anyone have any ideas?
Instead of looking at $a^{-1}$ and $b^{-1}$, try taking $e,\ a,\ b,\ ab,\ ba$ as your five elements, and show that they are all distinct.

5. Thanks, that worked beautifully!