4

I want to prove that the Groups of order upto 4 are abelian. I am not allowed to use concept of order of an element of a group.

Let $\displaystyle G=\{e,a,b,c\}$ be a group of order $\displaystyle 4$.

By closure, $\displaystyle ab\in G$. Clearly $\displaystyle ab\neq a,b$ ( as $\displaystyle ab=a\Longrightarrow b=e$ and similarly $\displaystyle ab\neq b$) so either $\displaystyle ab=e$ or $\displaystyle ab=c$.

Case 1: $\displaystyle ab=e$

It follows that $\displaystyle a^{-1} =b$ and thus our $\displaystyle G$ becomes $\displaystyle \{e,a,a^{-1} ,c\}$. Again by closure $\displaystyle ac\in G\Longrightarrow ac\neq a,c,e$. Since $\displaystyle ac=e\Longrightarrow a^{-1} =c$, which is not possible as $\displaystyle a^{-1}$ and $\displaystyle c$ are distinct.

So we must have $\displaystyle ac=a^{-1} \Longrightarrow c=a^{-2}$. Thus our $\displaystyle G$ finally becomes $\displaystyle \{e,a,a^{-1} ,a^{-2} \}$, which is clearly abelian.

Case 2: $\displaystyle ab=c$ so that our $\displaystyle G$ becomes $\displaystyle \{e,a,b,ab\}$.

Clearly, $\displaystyle a^{2} \neq a,ab$. So either $\displaystyle a^{2} =b$ or $\displaystyle a^{2} =e$. If $\displaystyle a^{2} =b$, then $\displaystyle G$ becomes $\displaystyle \{e,a,a^{2} ,a^{3} \}$, which is abelian.

And if $\displaystyle a^{2} =e$, then $\displaystyle ab^{2} \neq ab,b$ so either $\displaystyle ab^{2} =e$ or $\displaystyle a$. If $\displaystyle ab^{2} =e$ then $\displaystyle a^{2} b^{2} =a=b^{2}$, and the group becomes $\displaystyle \{e,b^{2} ,b,b^{3} \}$, which is abelian.

And if $\displaystyle ab^{2} =a$ then $\displaystyle b^{2} =e$.

$\displaystyle ( ab)( ba) =ab^{2} a=aea=e\Longrightarrow ( ab)^{-1} =ba$. $\displaystyle ba\neq a,b$ and $\displaystyle ba\neq e$ as $\displaystyle ba=e\Longrightarrow ab( ba) =ab=e$, which is not possible. So $\displaystyle ba=ab$. Therefore, $\displaystyle G=\{e,a,b,ab\}$ is abelian.

Groups of order $1,2$ and $3$ are trivially abelian.

Is my proof correct? Thanks.

Shaun
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Koro
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  • 2
    Looks right, though I would probably phrase it somewhat differently at places. – Arturo Magidin May 08 '21 at 20:05
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    It's correct, but you can short cut the proof a bit. Once you know that $ab=c$ (in Case $2$), reverse the roles of $a$ and $b$ to see immediately that also $ba=c$, so that $ab=ba$. – Robert Shore May 08 '21 at 20:07
  • I have made one correction in the proof for the case $ab^2=e$. – Koro May 08 '21 at 20:12
  • @ArturoMagidin: Thanks a lot for your review sir. – Koro May 08 '21 at 20:12
  • @RobertShore: Thanks a lot for your review sir. Reversing the roles of $a,b$ felt dangerous as $ab$ could be equal to something for example and what guarantees that $ba$ will also be that something since we apriori know nothing about $G$. But I think that you are right :) – Koro May 08 '21 at 20:14
  • Put differently, if $ab\ne ba$, then the five elements $a,b,1,ab,ba$ must be distinct: If $a=b$, then they commute. From $a1=1a$, w ehave $b\ne 1$, similarly $a\ne1$. Hence also $a\ne ab$ and $a\ne ba$ amd $b\ne ab$ and $b\ne ba$. If $ab=1$ (or $ba=1$) then $a,b$ are inversees and commute. And of course alsoe $ab\ne ba$. – Hagen von Eitzen May 08 '21 at 20:54
  • See https://math.stackexchange.com/questions/443642/prove-that-every-group-of-order-4-is-abelian – Nicky Hekster May 09 '21 at 17:47

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