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

I just learn abstract algebra by self study. I want help to solve this problem.
Just give me a hint.
 A: You need an instance of $ab\ne ba$. That requires $a\ne b$. Also $a\ne 1$ and $b\ne 1$ as $1$ commutes. Also, $a,b$ are not inverse of each other as those commute. Hence $1, a, b, ab, ba$ are pairwise distinct
A: In fact it must have at least $6$ elements.
You can discard the possibilities of the group having exactly $1$ element immediately. You can discard the possibility of the group having a prime number of elements because any such groups are cyclic, so $2,3$ and $5$ are discarded.
It remains to show that no non-abelian group with $4$ elements exists.
If it has an element of order $4$ then it is cyclic, otherwise every element must have order $2$ or $1$. And a group in which this happens is abelian, since $(ab)^2=e=a^2b^2$
A: Alternative solution: Suppose it is not abelian, then it has two elements $a$ and $b$ that do not commute, hence the group contains $e,a,b,ab,ba$ and must have at least $5$ elements.
A: Hint: Try to make a list of all groups of orders $1,2,3,4$ (up to isomorphism). There are not many (five to be precise) and you will see, they are all abelian. (One might add that there is also only one of order $5$ and it is also abelian.)
