# Prove that every group of order $4$ is abelian

How can I show that every group of order $4$ is abelian?

Let's denote $e,a,b,c$ as the four elements of the group.

Since $e$ is identity, we have $e*x=x*e$ for every $x$:

$$\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & & & \\ b & b & & & \\ c & c & & & \\\hline \end{array}$$

Now $a*a=$? We have several possibilities.

If I choose $a*a=b$ I can show $a*b=a*(a*a)=(a*a)*a=b*a$.

But so far I have shown only that $a$ and $b$ commute, there are also other pairs. And I have only discussed the choice $a*a=b$, there are also other possibilities. Is there a simpler way to do this?

Consider a group $G$ of order $4$. Suppose, towards a contradiction, that $G$ is not abelian. Then there must exist some distinct non-identity elements $x,y\in G$ such that $xy\ne yx$. But notice that:

• $xy≠e$ and $yx≠e$ (since $x$ and $y$ don’t commute, so $y≠x^{-1}$)
• $xy≠x$ and $yx≠x$ (since by hypothesis $y≠e$)
• $xy≠y$ and $yx≠y$ (since by hypothesis $x≠e$)

Thus, it follows that $e,x,y,xy,yx$ are $5$ distinct elements that are all in $G$. But this contradicts the fact that $G$ is of order $4$. Thus, $G$ must be abelian, as desired.

• +1 Yes, this directly shows that all groups of order $<5$ are abelian, simply because it takes so many distinct elements to merely formulate noncommutativity, so to speak. This is the very approach I usually present this fact. – Hagen von Eitzen Jul 14 '13 at 21:04
• Very simple and lucid. – Magneto Jul 18 '18 at 4:31

Let $G$ be a group of order $4$. If $G$ is cyclic, we're done. If not, $x^2 = e$ for all $x \in G$, which implies $(xy)(xy) = e$ for all $x, y \in G$. Multiply on the right by $yx$ to get $xy = yx$.

• You should remember to cite Lagrange's theorem and not take it for granted in this sort of situation. – Samuel Yusim Jun 19 '14 at 22:53

HINT: There are only two groups of order $4$. One of them has an element of order $4$; the other one doesn’t.

There are three possible scenarios:

• The group has an element of order 4

Let $x$ be the element of order $4$, then the group consists of $e,x,x^2, x^3$, which is commutative (actually cyclic)

• The group has an element of order 3

Let $x$ be the element of order $3$, then the group consists of $e,x,x^2$ and one more element $y$.

Then $xy$ has to be $e, x, x^2$ or $y$, and it is trivial to check that none of them is possible. This case is impossible.

• Every element has order at most 2

Then if $x,y \in G$ the elements $x, y$ and $xy$ have order 1 or 2. We have $x^2=y^2=(xy)^2=e$.

Then $$xxyy=x^2y^2=(xy)^2=xyxy \Rightarrow xy=yx$$

• Can the downvoter please explain what is wrong with this answer? – N. S. Nov 27 '15 at 13:02
• Could it be $x^2y^2=(xy)^2$ ? – zjk Mar 30 '18 at 15:51

We know that order of any element of a group divides the order of the group. So possible orders of elements of our are 1, 2, 4. Moreover, only identity has order equal to 1. So all other elements must have orders 2 or 4.

If there is an element of order 4, this group is cyclic.

So the only remaining case is that there might be a group with four elements where all non-identity elements are of order two.

We know that every group with this property is commutative, see Prove that if $g^2=e$ for all g in G then G is Abelian. or Order of nontrivial elements is 2 implies Abelian group

But for the case of 4 elements, we can also find this group by filling out the Cayley table.

Let us denote the elements of the group $G=\{e,a,b,c\}$, where $e$ is the identity. So far we know that $a^2=b^2=c^2=e$.

$$\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & & \\ b & b & & e & \\ c & c & & & e \\\hline \end{array}$$

Now we can continue in a sudoku-like fashion. From the cancelation law we know that we cannot have the same element twice in the same row/in the same column. (If an element appears twice in a row $x$, it means that $x*y=x*z$ for $y\ne z$. Cancelation law says that this cannot happen in a group.)

So, for example, $a*b$ equals neither $a$ nor $e$, since this would mean having repeated letter in a row $a$. But it cannot be equal to $b$ either, since $b$ already is in this column. So the only remaining possibility is $a*b=c$. Almost in the same way we get that $b*a=c$.

$$\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & c & \\ b & b & c & e & \\ c & c & & & e \\\hline \end{array}$$

Now we continue like this - if there is only one possibility, we can write down an element to this position. We can fill the whole table.

$$\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\\hline \end{array}$$

So we see that if a group on 4 elements is not cyclic, then it must have table like this. This table is symmetric w.r.t. the main diagonal, which means that the group is commutative.

Remark 1: So far we do not know whether the above table determines a group. We only know that if a non-cyclic group on 4 elements exists, it must have table like this. It can be shown that this is indeed a group. A very laborious way to do this would be checking associativity for all triples. But if we already know about the group $(\mathbb Z_2\times\mathbb Z_2,\times)$, it is not difficult to notice that it has "the same" table. $$\begin{array}{|c|cccc|} \hline & (0,0) & (0,1) & (1,0) & (1,1) \\\hline (0,0) & (0,0) & (0,1) & (1,0) & (1,1) \\ (0,1) & (0,1) & (0,0) & (1,1) & (1,0) \\ (1,0) & (1,0) & (1,1) & (0,0) & (0,1) \\ (1,1) & (1,1) & (1,0) & (0,1) & (0,0) \\\hline \end{array}$$

Remark 2: We have in fact proved something more that was asked. We have not only shown that every group on 4 elements is commutative. We have also shown that it is either cyclic or has the table described above. This means that there are only two groups having 4 element "up to isomorphism" (=all other groups have the same table, only elements are "renamed"). Namely, every group with $|G|=4$ is isomorphic either to $(\mathbb Z_4,+)$ or to $(\mathbb Z_2\times\mathbb Z_2,\times)$.

There are exactly two groups of order $4$: one is cyclic, and hence abelian, the other is not cyclic, but it is abelian. To see this, I'd suggest you get your hands "dirty" with the exercise below. Completing it will help you understand both the properties a group must satisfy, and also to learn about why all order-$4$ groups are necessarily isomorphic to one of two abelian groups of order $4$: $\mathbb Z_4$, and the Klein $4$-group.

Exercise: Try to complete the Cayley table for a group with elements $e, a, b, c$: one of which, say $e$, must be the identity. We know that every element must appear once and only once in each row and in column. Show that there are no more than, and no fewer than, two groups of order $4$, up to isomorphism. Confirm that each is abelian (the entries will be symmetric about the main diagonal of the $4\times 4$ Cayley table.)

• I think that we can take another way to prove this , which is , prove that any group of order less than $6$ is abelian , this was a theorem and its prove was given in a text called , selected topics in group theory . This is more general , hence stronger , result than we are asked!. – Fawzy Hegab Jul 16 '13 at 1:07
• Adriano way of proving " an above answer " uses the same methods in the proof of this theorem. – Fawzy Hegab Jul 16 '13 at 1:09
• Yes, indeed, @MathsLover. I think the OP is very new to group theory, so I was trying to suggest a method that the OP could use to get insight into the properties of a group, an abelian group, etc. Given that the task at hand (stated problem)...it seems the OP is first encountering the definitions of a group and an abelian group! – amWhy Jul 16 '13 at 1:14

Set $G = \{ e, a , b, c \}$ such that non of its elements has order 4 (else is isomorphic to ${\mathbb{Z}}_4$ and will became abelian) and also the elements are pairwise disjoint and not equal to the identity ; I claim that $ab = c$: because if $ab = e \Rightarrow a = b^{-1}.$ But by assumption $ord(b) = 2 \Rightarrow b^2 = 1 \Rightarrow b= b^{-1} = a$; that's contradiction. As well as if $ab = a \Rightarrow b=e$ and $ab = b \Rightarrow a=e$; that's contradiction. And by the same approach we can prove that $ba = c = ab$ and $ca = b = ac$ and $cb = a = bc$ and we are done.

• Note that $a \neq b \neq c \neq e$ does not exclude the possibility that $a = c$, $a = e$, or $b = e$. – Michael Albanese Nov 22 '15 at 22:09
• @ Michael Albanese I think now is correct. – Farrokh Nov 22 '15 at 22:26

There aren't as many choices as you'd think.

$(a, b, c)^4 = e$ so $(a^2, b^2, c^2)^2 = e$. So either $a^2 = e$ or $a^2 = b$ or $c$ and $b^2$ or $c^2 = e$. Whatever. Either way there is some non-identity element, x, is such that $x^2 = e$.

We might as well assume that element is $a$ (without loss of generality). So $a^2 = e$.

Now you can fill in the third row and column with:

$\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & & \\ c & c & b & & \\\hline \end{array}$

There's only 4 squares left. The only choice you have is $b^2 = a$ or $b^2 = e$. So basically there were only two options:

1: $\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\\hline \end{array}$

or

2: $\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & a & e \\ c & c & b & e & a \\\hline \end{array}$

Note:

$\#1 \cong \mathbb Z_4$ with $b = b; b^2 = a; b^3 = c; b^4 = e$

And $\#2 \cong \mathbb Z_2 \times \mathbb Z_2$ With $a^2 = b^2 = c^2 = e; a*b =c = b*a; b*c = a = c*b; c*a =b=a*c$.

That's it. Only two groups of order 4.

• It looks like #1 is actually the Klein group, isomorphic to Z2xZ2, and #2 is something I've just come across and can't understand. It's not Z2xZ2 and not Z4. – Ran Lottem Aug 10 '20 at 18:37