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

Question

Is it possible to prove that every group of order less or equal to five is abelian?

We know that groups of prime order are cyclic and therefore commutative. As the number $4$ is the only composite number $\le5$, it basically remains to show this for groups of order four.

Answer 1

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.

Answer 2

Yes, it is possible to prove. The question is, how much group theory you can use. Any group of prime order is cyclic, hence abelian. This implies that all groups of order $2$, $3$ and $5$ are abelian. On the other hand, all groups of order $p^2$ are abelian. Hence all groups of order $4$ are abelian. The trivial group of order $1$ is abelian as well. For groups of order $4$ the explicit classification also shows the result.

Answer 3

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

Answer 4

Another way to prove it is that a finite ordered non-abelian group must have order 6 or greater which can be done like this

if $|G|=1$ then it's trivial, So assume $|G|>1$ then there exist $a\in G$, but all groups have inverses so $a^{-1}\in G$, which means $|G|\geq 3$, but this case would be abelian so there must be another element $b\in G$ and equally so it's inverse, so we have that $|G|\geq 5$, But we also have that $ab\in G$, which would give one of the following $ab=a$, $ab=b$, $ab=e$, $ab=a^{-1}$ or $ab=b^{-1}$, the first ones can be eliminated as they imply that either $a$ or $b$ are identity or inverse, the latter two can be eliminated for the reason that they would make it abelian as we get $aba=a^2b=e$, with left and right a multiplication on $ab=a^{-1}$. This gives us that $ab=ba$ for the elements which clearly doesn't work as it is a non-abelian group and the remainder would be the same (with the inverses etc), so there must exist some other element then. So $|G|\geq 6$

Answer 5

$xy \neq yx \implies G$ 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.