Ryan Rueger

Groups of Order 4 are Commutative

March 30, 2021

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 you need at least 5 elements to even formulate the concept of non-commutativity.

Proof. Suppose $G$ is a group and $a, b$ elements thereof which do not 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 $5$.

Using this proof technique we can further prove that

any group of order 5 is commutative

Proof. Suppose there is a group $G$ of order 5 that is not commutative. Then there are two elements $a, b$ in $G$ that do not commute and we conclude that $G$ must be of the form $\{1, a, b, ab, ba\}$. Now consider the element $aba^{-1}$ in $G$. We note that $aba^{-1} \neq 1$ otherwise, $b = 1$; we also note that $aba^{-1} \neq a$ otherwise $b = a$; also $aba^{-1} \neq b$ otherwise $ab = ba$; and finally $aba^{-1} \neq ab$ otherwise $a^{-1} = 1 = a$. So the only remaining option is that $aba^{-1} = ba$, or equivalently $ab = ba^2$. By symmetry $ba = ab^2$. Inserting the first equation into the second yields $ba = ba^2b$ which is equivalent to $1 = ab = ba$. This is a contradiction to $a, b$ not commuting.

I called this the “naive” proof, because I think most mathematicians would instinctively try to use Lagrange’s theorem.

Theorem (Lagrange). Let $H$ be 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, $=_H$ is

(Reflexive) $a a^{-1} = 1 \in H$ hence $a =_H a$.
(Symmetric) $a =_H b$ implies $ab^{-1} \in H$. So $ba^{-1} = {(ab^{-1})}^{-1} \in H$ and $b =_H a$.
(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 over all cosets is $G$

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

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 $|H|$.

Or more specifically, most mathematicians would probably instictively try to use the following

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 $g$) 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

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

This covers the cases 2, 3, 5. Only the case of order 4 is missing. Maybe we can modify Lagrange’s Theorem to show that groups of order 4 are also commutative?

Indeed, there is the following

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 generalises the statement “groups of order 4 are commutative”.

Indeed, 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. If not, the order of all elements is at most 2, and thus the group is commutative by the Lemma.

Finally, we conclude

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 symmetric group on three letters, comprising the set of all permutations on a group of three things. One way to write this is as follows

$$ \{(), (ab), (ac), (cb), (abc) = (ab)(bc), (acb) = (ac)(cb)\} $$

where the notation $()$ denotes the empty permutation (i.e. the identity); $(xy)$ denotes the permutation of $x$ and $y$; and $(xy)(zw)$ denotes the compositions of two permutations.