# Show that every group G with |G|<6 is abelian.

Show that every group G with |G|<6 is abelian.

So I tried to prove this one by cases.

Case 1: Suppose |G| = 1, then G is the trivial group, and is abelian.

Case 2: |G| is prime, then |G| = 2, 3, or 5. Every group of order p, where p is prime is cyclic. Let g be a generator of any cyclic group G and let $a,b\in G$. Then, $\exists x,y \in \mathbb{Z}$ such that $a=g^x$ and $b=g^y$. Then we have that $$ab = g^x g^y = g^{x+y} = g^{y+x} = g^y g^x = ba$$ Hence, all cyclic groups are abelian.

Case 3: |G| = 4

This one was kind of hard to wrap my head around until my professor told me that only two groups have order 4, $\mathbb{Z}_4$ and $\mathbb{V}$, the Klein group. So my proof of this case was sort of brute force, showing that each group was abelian.

What if I was asked to prove that all groups of order less than 100 (Note that I understand this is not true)? Brute force would not be the most efficient, as there are many groups of order less than 100. Is there a more efficient way of proving case 3?

• For the groups of order |G|<100 you should just exhibit one non abelian group of order <100, disproving the claim. – Laurent Hayez Apr 2 '15 at 4:19

For the case when $|G| = 4$, there are two possibilities. If $G$ has an element of order $4$, it must be cyclic, hence abelian.
If $G$ doesn't have any elements of order $4$, then every non-identity element must have order $2$. Then picking any two non-identity elements, we know that $(ab)^2 = abab = e$ (since all non-identity elements have order $2$). Right-multiplying by $b^{-1}a^{-1} = ba$, we have
so that two generic non-identity elements commute, hence $G$ is abelian.