# Abelian groups and the order of the group

How can I prove that the groups with order less than 6 are Abelian.?

And the first non Abelian group is of the order 6.

• Depends on what facts you are allowed to use. Do you know that groups of prime order are cyclic, and cyclic groups are abelian, and groups of order square of a prime are abelian? Commented Mar 20, 2015 at 6:13
• So that means 2,3,5 are out and 4 is Abelian.. Yeaa gotit.... Commented Mar 20, 2015 at 6:14

To add to @Yval Filmus' answer, every group of prime order must be abelian, leaving all groups of order 4. Since $S_3$ is of order $3! = 6$ and is not abelian, a group of order $6$ is the smallest order for which a group is abelian.
Since $2$, $3$, and $5$ are prime, this leaves groups of order $1$ and $4$. A group of order $1$ is trivial, and there are only two groups of order $4$. Every group of order $4$ is isomorphic to either $\mathbb{Z}_{4}$ or $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$.
$\mathbb{Z}_{4}$ is abelian, and $G \oplus H$ is abelian if and only if $G$ and $H$ are abelian, thus $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ must be abelian since $\mathbb{Z}_{2}$ is abelian.
• In fact we don't need to prove that any group of order $4$ is isomorphic to either $\mathbb Z_4$ or $\mathbb Z_2 \oplus \mathbb Z_2.$ If a group of order $4$ has an element of order $4,$ then it is cyclic. If it has no element of order $4,$ then every element $g$ in that group satisfies $g^2=e.$ Any group with this condition is abelian. Commented Mar 20, 2015 at 7:08
• Right! Only one element of a group can have order $1$ (the identity). This means that all elements must have either order $2$ or $4$ by Lagrange's theorem (all subgroups must divide the order of the group, which is $4$). If there is an element of order $4$, then the group is cyclic, and cyclic groups are necessarily abelian. If there is no element of order $4$, then all elements must be of order $2$, which means the group is necessarily abelian. Thanks for pointing that out :) Commented Mar 20, 2015 at 7:18
• Yes! Your answer is absolutely correct. I merely pointed out that one doesn't need the isomorphisms for degree $4$ as it may create some unnecessary complication. Since we need only to prove that it is abelian, we can avoid the isomorphisms. +1 Commented Mar 20, 2015 at 7:25