# A fun problem from Charles C. Pinter’s “A Book of Abstract Algebra.”

Back when I was in college, I took a course on Modern Algebra. It was terribly interesting, but I was also taking four other courses at the time (Integrated Circuit Design, Electromagnetic Theory, Complex Analysis, and an electronics lab course), so I wasn’t really able to digest the subject as well as I’d have liked.

In my spare time recently I’ve been powering through Charles C. Pinter’s “A Book of Abstract Algebra, Second Edition” and I came across this fun problem in the chapter on subgroups. The problem involves an interestingly-chosen subset of a group… it is similar to the group’s Center, but not quite the same.

The problem is as follows: “Let $G$ be a group and let $C' = \left\{a \in G : (ax)^2 = (xa)^2 \text{\ for every\ } x \in G \right\}$. Prove that $C'$ is a subgroup of $G$.”

I admit this one took me a little while, and I did have to ask for a hint. But in the end I got it, and I’m going to share the solution (and the line of thought behind it) here.

To prove that $C'$ is a subgroup of $G$, we need to show two things:

(i) $a,b \in C' \Rightarrow ab \in C'$. That is, $C'$ is closed under products.
(ii) $a \in C' \Rightarrow a^{-1} \in C' \text{\ and\ } aa^{-1} = a^{-1}a = e \in C'$. That is, that $C'$ is closed under inverses.

For (i), we know that $G$ is a group, and $C' \subset G$, so $\forall a, b \in C', \forall x \in G, bx \in G$. Thus by the definition of $C'$, we know $(a(bx))^{2} = a(bx)a(bx) = (bx)a(bx)a = ((bx)a)^{2}$. The second equality here is the critical step, where we must invoke the definition of $C'$ to swap the order of $a$ and $bx$. Let’s call this relation [1].

But hold on! $b$ was also in $C'$, wasn’t it? And $xa \in G$ too. So we also know that $(b(xa))^{2} = b(xa)b(xa) = (xa)b(xa)b = ((xa)b)^{2}$. Let’s call this relation [2].

Since G is a group, associativity holds (we already used it above, but it should be emphasized here) and the right hand side of [1] is equal to the left hand side of [2]. Thus we can make the following equation with the assistance of good old transitivity: $(ab)x(ab)x = a(bx)a(bx) = (bx)a(bx)a = b(xa)b(xa) = (xa)b(xa)b = x(ab)x(ab)$. Therefore $a, b \in C' \Rightarrow ab \in C'$.

For (ii), we know $a^{-1} \in G$ since $G$ is a group. Therefore $\forall a \in C', a^{-1}xa^{-1} \in G$ and therefore by the definition of $C'$, we know that $\left(\left(a^{-1}xa^{-1}\right)a\right)^{2} = \left(a\left(a^{-1}xa^{-1}\right)\right)^{2}$.

By associativity, the left hand side of the above is equal to $\left(a^{-1}x\right)^{2}$, and the right hand side is equal to $\left(xa^{-1}\right)^{2}$.

So we now have $\left(a^{-1} x \right)^{2} = \left( x a^{-1} \right)^{2}$. The identity element of $G$, $e = a^{-1}a$ is “obviously” in $C'$ also as $(ex)^2 = x^2 = (xe)^2$. Therefore $C'$ is closed under inverses. Since (i) and (ii) hold, $C'$ is a subgroup of $G$. QED.

Whew. That wasn’t so bad, was it?

Stay tuned for more fun math adventures!