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Suppose you have $(G, \cdot)$a group whose all elements are either of order $1$ or $2$. Show that it's abelian.

My proof, suppose you have $a,b \in G$ such that $a^1 = e$ and $b^1 =e$ then:
$$a \cdot b= e \cdot e = b \cdot a$$

Now suppose we have $a^1 = e$ and $b^2 = e$:
$$a \cdot b = e \cdot b = b \cdot e = b \cdot a$$

And finally suppose $a^2 = e$ and $b^2 = e$ then:
$$a \cdot b = a \cdot ( a^{-1} \cdot a ) \cdot b = (a^{-1})(a^2)b=a^{-1}b$$ and $$ b \cdot a = b (b^{-1} \cdot b ) a = b^{-1}(b^2)a = b^{-1}a$$ Thus we have $$(a\cdot b)(b\cdot a)= (a^{-1} \cdot b)(b^{-1} \cdot a)= e$$ Thus $ab=ba$.

Is my proof correct?

John Mayne
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2 Answers2

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Your first two cases are a little inelegant, since $a^1 = e$ means that $a=e$. So the property that every element is of order 1 or 2 can be simplified to requiring $$x^2 = e,\; \forall x \in G.$$

Here is a streamlined version. Let $a, b \in G$ and let $x = ab$. Since $x^2 = e$, we have $$ abab = e $$ Multiply on the right by $b$ and remember that $b^2 =e$ to get $$ aba = b $$ Multiply on the right by $a$ and remember that $a^2 =e$ to get $$ ab = ba $$

amWhy
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Matthew Leingang
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    Did you mean to write that $x^2 = x,; \forall x\in G;$ is a simplification of the property that every element in G is of order 1 or 2? Wouldn't we rather have that $x^2 = e, ; \forall x \in G;? – amWhy Sep 21 '17 at 18:04
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    Take the abelian group the Klein-IV Group: $G = {e, a, b, c}.;$ $e = e^2 = a^2 = b^2 = c^2$, but $a^2 \neq a, b^2 \neq b, c^2 \neq c...$. We also know that $a\cdot b = c, a\cdot c = b, b \cdot c = a$. So perhaps you can understand that I would question your second sentence in your answer. – amWhy Sep 21 '17 at 18:11
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    Anyway, I corrected what I take to have been a typo...(My two comments above are addressing your claim that "every element is of order 1 or 2 can be simplified to $x^2 = x,$ for all $x\in G$.) – amWhy Sep 21 '17 at 18:19
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    $a^2 = a \Longleftrightarrow a = e$! – Robert Lewis Sep 21 '17 at 18:19
  • @amWhy: Right you are. Sorry for my typo. I was distracted by work and didn't keep track of my answer! – Matthew Leingang Sep 21 '17 at 18:57
  • No problem... It was clear to me, further on in your proof that that's what you meant (I.e., originally just a typo; Believe me, I've had my share of typos!). I typically don't edit another's answer, but I also remember the times when I wish someone would have fixed one of my typos! No worries; your answer is fine! – amWhy Sep 21 '17 at 19:25
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It appears that OP John Mayne's questions about his attempt have been sufficiently addressed by other users; the key, I think, is to realize that $g^2 = e \Longleftrightarrow g = g^{-1}$, viz:

The condition

$\forall g \in G (g^1 = e \vee g^2 = e) \tag 1$

is logically equivalent to

$\forall g \in G (g^2 = e), \tag 2$

since

$(g^1 = e \Longleftrightarrow g = e) \tag 3$

and

$g = e \Longrightarrow g^2 = e; \tag 4$

therefore we need to show

$\forall g \in G (g^2 = e) \Longrightarrow \forall a, b \in G (ab = ba). \tag 5$

Now

$\forall g \in G (g^2 = e \Longleftrightarrow g^{-1}g^2 = g^{-1}e \Longleftrightarrow g = g^{-1}); \tag 6$

thus

$\forall a, b \in G (ab = a^{-1}b^{-1} = (ba)^{-1} = ba). \tag 7$

Robert Lewis
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