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This is the next problem fron Herstein's book:

(a) If the group $G$ has three elements, show it must be abelian.

(b) Do part (a) if $G$ has four elements.

My proof:

a) Since $|G|=3$ then $G=\{a,b,e\}$ and since $G$ is closed under group operation then $ab\in G$ and $ab\notin\{a,b\}$. Hence $ab=e$ and $ba=e$ and we get that $G$ is abelian group.

b) Since $|G|=4$ then $G=\{a,b,c,e\}$ and $ab\in G$ and it's easy to see that $ab\notin \{a,b\}$. The case $ab=e$ gives us the identity $ba=e$ and therefore $ab=ba$.

We need to consider the case $ab=c$. In this case $ba\in \{c,e\}$.

  1. If $ab=c$ and $ba=e$ then we derive $c=e$ which is contradiction.
  2. If $ab=c$ and $ba=c$ we get that $ab=ba$ and we have done.

The same reasoning can be applied for pairs $(a,c)$ and $(b,c)$.

Is above reasoning correct?

If any can suggest more shorter and elegant proof it would be great.

RFZ
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  • You can see here – S.S Nov 19 '17 at 18:28
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    Having said that, I don't see anything wrong with your proof. –  Nov 19 '17 at 18:30
  • Dear RFZ, for "more shorter and elegant proofs" you need to learn Lagrange, and cyclic groups, see here. Then it follows immediately that a group of prime order is abelian. For order $4$ there are already many good solutions, see above. – Dietrich Burde Nov 19 '17 at 19:05

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