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This question already has an answer here:

What is the intuition behind the fact that if every element in a group is of order $2$, we have that the group is abelian? I can prove it, but I do not know the intuition behind it.

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marked as duplicate by Namaste abstract-algebra Sep 18 '16 at 0:09

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    $\begingroup$ Perhaps that there are simply not enough "degrees of freedom" to be nonabelian? You just can't get any nonabelian relations once you have that $aabb=abab=e$ for all$a,b$. $\endgroup$ – user452 May 30 '14 at 0:24
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    $\begingroup$ I mean the availability of relations on the group elements. That is, equations the group elements satisfy. You immediately get the relation I provided above, and it instantly rules out any nonabelian relation. $\endgroup$ – user452 May 30 '14 at 0:31
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    $\begingroup$ I don't think there's much of an intuition. I don't see how an intuitive explanation could distinguish $2$ from $3$ without just going through the proof, but the conclusion is false with $2$ replaced by $3$. $\endgroup$ – Qiaochu Yuan May 30 '14 at 0:54
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    $\begingroup$ A small comment: no group has the property that its every element has order $2$, since the identity always has order $1$. $\endgroup$ – goblin May 30 '14 at 1:18
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    $\begingroup$ @user: it's common to see people say "order $n$" when they mean "$n$-torsion," e.g. order dividing $n$. I think the meaning is clear. $\endgroup$ – Qiaochu Yuan May 30 '14 at 5:10
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I think it is a special case of a general case which is;

Lemma :if $f:G\to G$ by $f(x)=x^{-1}$ is a homomorphism if and only if $G$ is abelian.

Now think that you have this lemma and you are looking for a case which can happen. The most natural idea is setting; $x=x^{-1}$ for all $x$ i.e $x^2=e$ then $f$ becomes identity map which is a homomorphism so we should have an abelian group.

Now, you can ask that what is the intuation of the lemma ? It is also natural since every element has uniqe inverse then the map is a bijection and it is natural to ask when this bijection is a homomorphism.

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    $\begingroup$ I think its cool that if the inversion function $G \rightarrow G$ is a homomorphism, then (by your lemma) so too is the law of composition $G^2 \rightarrow G.$ $\endgroup$ – goblin May 30 '14 at 2:22
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I think of it like this: If $a,b$ are in the group, then $ab$ is in the group and of order 2 by assumption... but this means

$$ (ab)^2=abab=e \implies ab=ba $$

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    $\begingroup$ Nice answer. A small comment: it would be more correct to say that $ab$ is of order at most $2$ with respect to the divisibility order. $\endgroup$ – goblin May 30 '14 at 1:19
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I'm not exactly sure what you're looking for in an intuitive explanation, so I'll just explain how I look at it. Hopefully it will be helpful for you as well.

The statement that every element has order 2 is equivalent to the statement that every element is its own inverse. If you are comfortable with inverses being unique, then commutativity is simply the observation that $abba$ is the identity for any $a$ and $b$ in the group, because it shows that $ba$ is the inverse of $ab$. Does this make it any more clear?

Edit: Mesel's answer inspired me to think on this a little more. In any group you have a composition operation and inverses. So we can take two elements of the group $a$ and $b$ (with inverses $a^{-1}$ and $b^{-1}$) and combine them to get $ab$. A natural question to ask is what the inverse of this new element $ab$ is, in terms of the four elements we started with. It's not hard to see that it is $b^{-1}a^{-1}$.

Now we have this interesting relation $(ab)^{-1}=b^{-1}a^{-1}$ in which the elements are flipped. This suggests that there might be some connection between inversion and commutativity. The lemma that mesel mentioned describes that connection precisely (and as a bonus is easy to prove).

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Perhaps it would help to think in terms of permutations (at least for finite groups). Let $G$ be a group where every elements has order at most 2, and let $\sigma, \tau \in G$ have order 2. Then $\sigma$ and $\tau$ are both products of cycles of length 1 or 2, and furthermore so are $\sigma\tau$ and $\tau\sigma$.

Now think about the 2-cycles of $\sigma$: suppose that one of them is $(a\ b)$. Then $\tau$ has either $(a\ b)$ or $(a)\ (b)$. For if not, then, without loss of generality, $\tau$ has a 2-cycle of the form $(a\ c)$, but in this case you can see that $\sigma\tau$ and $\tau\sigma$ will not have order 2.

In other words, each 2-cycle in $\sigma$ either matches up and cancels out with a corresponding 2-cycle in $\tau$, or is left alone by $\tau$. You can then see that the order in which you apply $\sigma$ and $\tau$ clearly doesn't matter.

I don't know if this is short enough for an intuition, but I hope it helps.

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Clarifying Jeb's answer a little:

$(ab)^2 = e\\ abab = e$

Multiply by $a$ on the left:

$aabab = a\\ (aa)bab = a\\ bab = a$

Multiply by $b$ on the right:

$babb = ab\\ ba(bb) = ab\\ ba = ab$

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