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?
Best Answer
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
\begin{align*}abab(ba) &= e(ba)\\ ab &= ba, \end{align*}
so that two generic non-identity elements commute, hence $G$ is abelian.
Eventually, you'll learn the fundamental theorem of finite abelian groups (ooh, fancy!) to efficiently classify all finite abelian groups.