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Here is Prob. 9, Sec. 3.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:

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

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

(c) Do part (a) if $G$ has five elements.

I think I am clear on how to tackle Part (a).

So here I will only be attempting Part (b).

My Attempt:

Suppose our group $G$ has four distinct elements, say, $e, a, b, c$, with $e$ being the identity element.

Now suppose, if possible, that $ab \neq ba$.

As the elements $e, a, b, c$ of $G$ are all distinct, so by virtue of the cancellation laws (i.e. Lemma 2.3.2 in Herstein), we cannot have $ab = a$, $ab = b$, $ba = a$, or $ba = b$.

Therefore we must $ab, ba \in \{ e, c \}$.

Since $ab \neq ba$ according to our supposition, there we can assume without any loss of generality that $ab = c$ and $ba = e$.

Thus our group $G$ has the four distinct elements $e, a, b, ab$.

Since $ba = e$, we also have $$ a = ae = a(ba). \tag{1} $$

However, since $G$ is a group, we must have $$ a(ba) = (ab)a. \tag{2} $$

From (1) and (2) above, we obtain $$ a = (ab)a, $$ from which we obtain $$ e = ab, $$ again by Lemma 2.3.2 in Herstein. This contradicts the fact that $e$ and $ab$ are two distinct elements of $G$. So our supposition that $ab \neq ba$ is wrong. Therefore we must have $$ ab = ba. $$

An analogous argument yield $bc = cb$ and also $ca = ac$.

Thus any two of the elements $a, b, c$ of $G$ commute. And, the identity element $e$ of course commutes with itself as well as with each of $a$, $b$, and $c$.

Hence our group $G$ must be abelian.

Is this proof correct? If so, is it rigorous enough for Herstein? Or, are there problems and issues?

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Yes your proof in fact is correct. But here are a few possible shorthands. First you only need to work with one non commuting pair of elements:

Suppose $G$ is not abelian, then there is some pair $a,b\in G$ such that: $ab\ne ba$. (We then can conclude that $a,b\ne e$). As you then found out: $ab,ba\in\{e,c\}$ for the fourth element $c\ne e,a,b$. So either $ab=e$ or $ab=c$ resulting in $ba=e$ (otherwise $a$ and $b$ commute and we are done). So in particular $a$ and $b$ are inverse elements to one another and hence commute. A contradiction. $G$ is abelian!

But you proof is just fine! :) Also nice effort of writing it down!

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