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I recently came across two properties that imply that a group is abelian:

  • If the order of each element except the identity element of a group is two then the group is abelian.
  • If every element of the group is its inverse then the group is abelian.

Obviously this comes from my module with no proof but I also looked in my book and found none of this there either, so I am interested in knowing how we can prove it?

Thanks.

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These are not properties of abelian groups, but properties which are sufficient to have an abelian group. They are not necessary. –  Raskolnikov Feb 12 '11 at 12:58
    
@Raskolnikov:So if any of this properties holds then the group is abelian under any binary composition? –  Quixotic Feb 12 '11 at 13:00
    
@Debanjan: are you really asking whether in all abelian groups the order of every element is 2? –  lhf Feb 12 '11 at 13:10
2  
For both: write $1 = (ab)(ab)$ and multiply with $ba$ from the right or left, as you prefer. –  t.b. Feb 12 '11 at 13:14
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It's funny if you write it like this: $ab = a(ab)(ab)b = (aa)ba(bb) = ba$. –  Myself Feb 12 '11 at 20:23

4 Answers 4

up vote 3 down vote accepted

In fact, the following is a stronger condition:

Prop. Let $G$ be a group, and let $x$ and $y$ be elements of $G$. If $$(xy)^2 = x^2y^2$$ then $x$ commutes with $y$; that is, $xy=yx$.

If every element of the group has order dividing $2$, then both sides of the equation above are equal to $1$, so the equality holds for any two elements of $G$; if every element is its own inverse, then every element has order $1$ or $2$, then again both sides equal the identity, so the result follows for every pair of elements. If any pair of elements commute, then the group is abelian.

This is a stronger result, though, because it also applies in other cases, when neither side is the identity but they are nonetheless equal.

Proof. Write the equation explicitly as $$xyxy = xxyy.$$ Multiplying on the left by $x^{-1}$ and on the right by $y^{-1}$, We get $$x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1}$$ which is the same as $$yx = xy.$$ So $x$ and $y$ commute. $\Box$

Now, try proving the following:

Proposition. Let $G$ be a group, and let $x$ and $y$ be two elements of $G$. The following are equivalent (that is, if one of them holds, then they all hold; if any on of them is false, then they are all false):

  1. $xy=yx$.
  2. $(xy)^2 = x^2y^2$.
  3. $(xy)^{-1} = x^{-1}y^{-1}$.
  4. There exists an integer $n$ such that $$(xy)^{n+i} = x^{n+i}y^{n+i}$$ holds for $i=0,1,2$.
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Suppose $x^2=1$ for all $x$ in the group. This is equivalent to $x=x^{-1}$ for all $x$. Now consider $x=ab$.

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By converse I assume you mean

$G$ abelian $\Rightarrow$ the order of $g \in G$ is two

Counterexample:

Let $G$ be $\mathbb{Z} / 3 \mathbb{Z}$ with addition mod 3, i.e. $G = \{ 0, 1, 2\}$. This group is clearly abelian but $1 + 1 \neq 0$, so the order of $1$ is not $2$, so the converse doesn't hold.

I assume you are aware that both statements you gave are equivalent.

Here is a proof of the claim:

claim: $\forall g \in G: \hspace{3mm} gg = e \hspace{3mm}$ then $G$ is abelian

proof:

Let $a$,$b$ $\in G$. Then $a = a ^{-1}$ and $b = b^{-1}$. Let $e$ be the neutral element of $G$. We want to show

$$ ab = ba$$

Because $G$ is a group we know if $a,b \in G$ then $ ab \in G$ and by assumption $abab = e$. But

$$ e = abab = ab a^{-1}b^{-1} = ab (ba)^{-1}$$

Multiplying both sides with $ba$ yields

$$ ba = ab$$

Note that I have used $(ab)^{-1} = b^{-1}a^{-1}$ which holds for any group and you have to prove as well.

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Look at this if it helps:

  • Since all your elements are of order $2$ you have $x^{2}=e$ for all $x \in G$ which implies $x=x^{-1}$. If you have two elements $a,b \in G$, then by above you have $a \cdot b = a^{-1} \cdot b^{-1}=(ba)^{-1}=ba$. Hence your group is abelian.

  • You have $x =x^{-1}$ for all $x \in G$. Continue doing as I proceeded for the first one. You should be able to get it.

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