## Every group of order 5 is abelian

Today I was reading some basic group theory from Herstein’s Topics in Algebra, and saw the following cute problem: Prove that every group of order 5 is abelian. Now, clearly Lagrange’s theorem implies that there is only one group of order 5, the cyclic group of order 5, which is obviously abelian. It would be too much cheating to resort to a such theorem, because Lagrange’s theorem is introduced later in the book. This gives me the impression that Herstein wants the reader to solve this problem by using only the definition of a group, and its basic properties (i.e. uniqueness of inverse element and etc.). I must admit I had a lot of fun coming up with a solution, but I feel like this is not the intended solution, because it is far from being an elegant argument.

Proposition. Every group of order 5 is abelian.

Proof: Assume a group $G$ has 5 elements. To be more concrete, let $G=\{a,b,c,d,e\}$ where $e$ is the identity element of $G$. We will assume, to the contrary, that $G$ is not abelian. Then, there exists two elements $x,y\in G$ such that $xy\ne yx$. But clearly, for this statement to be true, $x\ne e$ and $y\ne e$. Thus, we will assume, without loss of generality that $x=a$ and $y=b$. In other words, we assume $ab\ne ba$.

Now, observe that $ab\ne a$, $ba\ne a$ (because both implies $b=e$, which is false). Similarly, $ab\ne b$ and $ba\ne b$ (because both implies $a=e$ which is false). Also, we claim that $ab\ne e$. Because otherwise, $a$ would be inverse of $b$ which would imply that $ab=e=ba$, contrary to our assumption that $ab\ne ba$. Similarly, $ba\ne e$. Thus, $ab,ba\in \{c,d\}$. Without of loss of generality, we will assume $ab=c$ and $ba=d$.

The next step is to observe that $db=(ba)b=b(ab)=bc$ and $ad=a(ba)=(ab)a=ca$. We have shown that $db=bc$ and $ad=ca$. The cancellation law shows that $db=bc\ne b,c,d$. Also, we claim that $db=bc\ne e$, because otherwise we would have two elements $c$ and $d$ that serve as inverse for element $b$. But then, by uniqueness of an inverse element, it follows that $c=d$, contradiction! This forces $db=bc=a$. Similar analysis shows $ad=ca\ne a,c,d,e$, thus $ad=ca=b$.

Let’s see what we have. $bc=a$ implies $bab=a$. Multiply both sides by $a$ from the right to get $(ba)^2=a^2$. Similarly, $ca=b$ implies $aba=b$. Multiply both sides by $b$ from the left to get $(ba)^2=b^2$. This implies $a^2=b^2$ (because both are equal to $(ba)^2)$.

Finally, we can analyze $a^2=b^2$. Clearly, $a^2=b^2\ne a, b$ (because it would otherwise imply $a=e$ or $b=e$). Furthermore, $a^2\ne c$ because otherwise, $a^2=ab$, or $a=b$ which is false. Similarly, $a^2\ne d$ because otherwise $a^2=ba$, or $a=b$ which is false. This leaves us no other choice but $a^2=b^2=e$. But then, $cb=(ab)b=ab^2=a$. But recall that $db=bc=a$. In particular, $db=a$. Combining this with $cb=a$, we get $db=cb$. Cancelling $b$ from the right, we get $d=c$, which is a contradiction! This shows that having $ab\ne ba$ is impossible, and so $G$ must be abelian, completing the proof. $\blacksquare$

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