# Prob. 9 (b), Sec. 2.3, in Herstein's TOPICS IN ALGEBRA, 2nd ed: Every group of order $4$ is abelian

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?

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!