If $G=\{1, a, b, c\}$ is a group and $|g|<4 \forall g\in G$, show the group is abelian.

Question

I was working on this exercise from Dummit and Foote's abstract algebra today:

Exercise 1.1.36

Assume $G=\{1, a, b, c\}$ is a group of order 4 with identity 1. Assume also that $G$ has no elements of order 4 (so every element has order $\leq 3$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.

Now, I got stuck in my workings. Since it's an even ordered group we can assume $|a|=2$ (there must be such an element so at most we need a re-labeling). This gives us 2 options: $|b|=|c|=2$, making the group abelian. Or the option $|b|=3$, in which case $b^2=c$.

This is how far I got on my own, and I couldn't figure out how to continue from here on out, so I went ahead and searched for a solution manual, which gave the following response:

Exercise 1.1.36

We know that at least one element must be its own inverse from exercise 1.1.31. WLOG, assume that this element is $a$. From 1.1.31 we know that $b$ is its own inverse iff $c$ is its own inverse: if both are their own inverses, then the group is abelian by 1.1.25 and we are done.

If neither $b$ nor $c$ are their own inverses, then they must have order $3$ $(b^3 = c^3 = 1)$. We cannot have $b^2=b$ (for we would have $o(b) = 1$) or $b^2= 1$ (for we would have $o(b) = 2$) or $b^2=a$ (for we would have $b=b$, $b^2=a$, $b^3=ab$, $b^4=a^2= 1$ so that $o(b) = > 4$) so we must have $b^2=c$. And this gives us a contradiction, since it would imply that $o(b^2) =o(c) =o(b)$.

My problem with this proof is: Why is this last point a contradiction? If $c= b^2$, then $c^2 = b^4 = b$, $c^3 = b^6 = b^3b^3 = 1$. This seems really reasonable to me.

It'd be great if someone could explain this to me.

Answer 1

A relation $b^3=e$ in a group $G$ of order $4$ implies that $b=e$, which is a contradiction. Indeed, $b^3=e$ means either $o(b)=3$, which is impossible, because the order of an element must divide the order of the group, which is $4$. But $o(b)=2$ is also impossible, because of $b^2=e$ would imply $b^3=b\neq e$. So this is the contradiction in your proof.

Answer 2

The option $|b|=3$ can obviously be discarded because the group has order $4$.

Without resorting to Lagrange’s theorem, if the order of $b$ is $3$, then $b^3=1$ and $(b^2)^2=b$. Since $a$ has order $2$, the possibilities are $b^2=b$ or $b^2=c$. The former leads to $b=1$, a contradiction. Therefore $b^2=c$ and also $c$ has order $3$.

Thus the group is $\{1,a,b,b^2\}$. Note that $ab$ can only be $1$: from $ab=a$ we get $b=1$; from $ab=b$, $a=1$; from $ab=b^2$, $a=b$. However $ab=1$ implies $b=a^{-1}=a$. Contradiction.

We could also build the Cayley table; cancellation implies that every row and column cannot contain repeated elements. The top row and left column are just for naming, they don't belong to the table proper. \begin{array}{c|cccc} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 \\ b & b & \\ c & c & \end{array} The product $ab$ cannot be $b$, so it must be $c$; similarly, $ba=c$; this also force $ac=b$ and $ca=b$: \begin{array}{c|cccc} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 & c & b\\ b & b & c & \\ c & c & b & \end{array} If $b^2\ne1$, then it must be $a$ (because $c$ already appears in the same row). But this forces $b$ to have order $4$, because then $b^3=bb^2=ba=c\ne1$. \begin{array}{c|cccc} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 & c & b\\ b & b & c & 1 &\\ c & c & b & \end{array} Now the three empty slots are forced: \begin{array}{c|cccc} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 & c & b \\ b & b & c & 1 & a \\ c & c & b & a & 1 \end{array} Thus the operation is commutative.