A curious result on groups is that

*all groups of order 4 are commutative*.

I find it curious, because the cleanest and shortest proof of this statement is so elementary that it requires no other external results. In fact it is very insightful: it demonstrates that which demonstrates that a group must contain at least \( 5 \) elements to even formulate the concept of non-commutativity.

*Proof.*Suppose $(G, \cdot, 1)$ is a non-commutative group, with $a, b$ distinct (non-identity) elements thereof which don’t commute. Firstly, $ab \neq 1$, otherwise $ab = 1 = ba$ and $a, b$ commute. Moreover $ab \neq a, ba \neq a$, else $b$ is the identity; and by symmetry we know that $ab \neq b, ba \neq b$. So $G$ contains at least the distinct elements ${1, a, b, ab, ba}$. Hence $G$ must have order at least $5$.

When I was first
I called this the *“naive”* proof, because I think most mathematicians would try
to use Lagrange’s theorem, because it is used to show that all groups of *prime*
order are commutative. Maybe

**Theorem**(Lagrange).

*Suppose $H$ is a subgroup of the group $G$, then the order of $H$ divides the order of $G$.*

Define the equivalence relation $=_H$ on $G$ via

$$ a =_H b \overset{\mathrm{def.}}{\iff} ab^{-1} \in H $$

This is indeed an equivalence relation, $R$ is

- (Reflexive) $a a^{-1} = 1 \in H$ hence $a =_H a$.
- (Symmetric) $a =_H b$ implies $ab^{-1} \in H$. Hence $ba^{-1} = {(ab^{-1})}^{-1} \in H$. Hence $b =_H a$.
- (Transitive) If $a =_H b$ and $b =_H c$ then there exist elements $h_{ab}$ and $h_{bc}$ in $H$ which respecitvely are $ab^{-1} = h_{ab}$ and $bc^{-1} = h_{bc}$. Hence $ac^{-1} = (ab^{-1}) (bc^{-1}) = h_{ab} h_{bc}$ lies in $H$. So $a =_H c$.

With this, we can partition $G$ into cosets $aH = \{ah \mid h \in H\}$ for every $a$ in $G$. That is, the union

$$ \bigcup _{a \in G} aH = G $$

and for $aH \neq bH$ we have $aH \cap bH = \emptyset$. The latter claim is a direct consequence of the transitivity of $=_H$. So $G$ may be written as the disjoint union of cosets $aH$. These each are of size $|H|$.

A generalisation of *“all groups of order 4 are commutative”* is

**Lemma.**

*Groups whose elements have order at most 2 are commutative.*

*Proof*. If an element $g$ has order 1, it is the identity; if it has order
$2$, we have $g^2 = 1$ or equivalently $g = g^{-1}$. So in a group in which
all elements have order at most 2, we see that all elements satisfy $g =
g^{-1}$. Now, for any two non-identity elements $h, g$ with order 2, we have

$$ gh = g^{-1}h^{-1} = (hg)^{-1} = hg $$

with the last equality coming about because $hg$ is again an element of the group and as such satisfies $(hg)^{-1} = hg$.

This is indeed a generalisation of *“groups of order 4 are commutative”*
because, by Lagrange, the order of an element of a group of order $4$ must be
either 1, 2 or 4. If any element is of 4, the group is cyclic and thus
commutative. In all other cases, the order of all elements is at most 2, and
thus the group is commutative.

**Corollary**(of Lagrange).

*Groups of prime order are commutative.*

*Proof*. Let $G$ be a group of prime order. By Lagrange, the order of any
subgroup of $G$ must divide the order $|G|$ of the $G$. Since $|G|$ is prime,
the only subgroups are the trivial subgroup $\{1\}$ and the whole group.
Hence any element $g$ in $G$ either generates the whole group $G$ or the
trivial group. If $g$ is not the identity, it must generate a non-trivial
subgroup (since the subgroup generated by $g$ contains at least $g$ — a
non-identity element) and hence the whole group. Thus $G$ is cyclic, with
every element $h$ in $G$ representable as some power $g^k$ for $k$ an integer.
Clearly now, all elements commute, as the element $g$ commutes with itself.
That is, for $h = g^k$, and $h' = g^l$ in $G$ we have

$$h h' = g^k g^l = g^{k+l} = g^{l + k} = g^l g^k = h' h.$$

So in fact, we can conclude that

*all groups of order at most 5 are commutative*.

We showed above that on the one hand, a group needs at least 5 elements to be non-commutative, but also now that groups of order 5 are commutative because they have prime order.

Is there a group of order 6 that is not commutative? Sure. The group of symmetries of an equilateral triangle.

**Definition** (Symmetry). Given a set $F$ in a metric space $(M, d)$ a
*symmetry* $\varphi$ of $F$ is a distance preserving bijection $\varphi \colon F
\to F$. That is, for every pair of points $p, q$ in $F$, we have

$$d(p,q) = d(\varphi(p), \varphi(q))$$

The group of symmetries of an equilateral triangle consists of the identity $\mathrm{id}$; two non-trivial rotations of a third of a full turn $\varphi_1$ and two-thirds of a full turn $\varphi_2$; and three reflections $\psi_1, \psi_2, \psi_3$ each on a line passing through one of the corners of the triangle and perpendicular to the opposite edge.

Are we sure that these are all of the symmetries of the equilateral triangle? We observe that any symmetry must map each edge of the triangle to another edge. Hence there are at most $3! = 6$ different symmetries. We have accounted for all of them.

Is this group really not commutative? In general, two reflections do not commute. Consider what happens to $P_1$ when first applying $\psi_1$ and then $\psi_2$: $\psi_1$ will not move $P_1$ and $\psi_2$ will send it to where $P_3$ is intially. However, if we apply $\psi_2$ first, $P_1$ will move to where $P_3$ is intially, and then $\psi_1$ will move this to where $P_2$ was initially.