# Prove that every group of order $4$ is abelian [closed]

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

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

This is all I could proceed.

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## closed as off-topic by YACP, Amzoti, Danny Cheuk, tetori, AangJul 19 '13 at 4:22

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What are the possible orders of the elements? –  Daniel Fischer Jul 14 '13 at 20:44
Downvoted because 1. It shows no effort, 2. It looks like homework but is not tagged as such, and 3. Even a pure brute-force approach wouldn't take long. –  dfeuer Jul 14 '13 at 20:52

Every group of order 4 is isomorphic to Z4 or Klein V 4 group.

If it is isomorphic to Z4, we already know that Z4 is an abelian group, so it's not hard to prove it.

If it is isomorphic to Klevin V4 group, again it's not hard to prove it.

Therefore, its abelian.

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Indeed. this is very close to saying «This is true because it is true»! –  Mariano Suárez-Alvarez Jul 14 '13 at 21:07
–  Pedro Tamaroff Jul 14 '13 at 21:13

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.

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+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

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 $xy = yx$.

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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

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.)

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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!. –  Maths Lover Jul 16 '13 at 1:07
Adriano way of proving " an above answer " uses the same methods in the proof of this theorem. –  Maths Lover 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

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

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