Prove every group of order less or equal to five is abelian [closed]

Order of a group , $n$ starts from 1.

When $n=1$ the group is a trivial one.

Now every group of prime order is cyclic and hence abelian. Hence groups of $n=2,3$ and $5$ are abelian.

Since every group of order $\ p^2$ (where $p$ is prime) is abelian. Group of order $4=\ 2^2$ is abelian.

Hence every group of order less than or equal to $5$ is abelian.

Group of order 1 is trivial, groups of order 2,3,5 are cyclic by lagrange theorem so they are abelian.

For a group of order 4, if it has an element of order 4, it is abelian since it is cyclic(isomorphic to Z4).

If orders of every element are 2, then the inverse of an element is the element itself so you can verify every element commute with each other so the group is abelian(isomorphic to klein four group)

You can make the cayley table(operation table) for the group of order 4. You will find 4 possibilities. 3 of them are just same groups with different elements' name and are all isomorphic to Z4.

The other one is the klein four group. Those are all abelian if you check the table

$xy \neq yx \implies G$ has at least five elements: $e,x,y,xy,yx$. But $|G| = 5$ means that $G$ is isomorphic to $\mathbb{Z}_5$, so every group of order less than 6 must be abelian.