Identifying Inverse Groups

Question

Let $G$ be a group $(S,\cdot)$. Let’s define its inverse, $G^{-1}$, as $(S,/)$, where $a/b=a \cdot b^{-1}$. My question is, when are both $G$ and $G^{-1}$ groups? Right now the only groups I can think of are groups with the symmetric difference/XOR operation, because then $G=G^{-1}$. Are there more examples? For the finite case, if $G=G_1 \times G_2$ then $G^{-1}$ is a group iff $G_1^{-1}$ and $G_2^{-1}$ are both groups, so I guess this question can be simplified to infinite groups and finite simple groups.

Answer 1

I will stick to multiplicative notation. You have a group $(G,\cdot)$, and want to define $(G,\star)$, where $a\star b = ab^{-1}$.

If $\star$ yields a group, then in particular we have $$ab^{-1}c^{-1} = (ab^{-1})c^{-1} = (a\star b)\star c = a\star(b\star c) = a(bc^{-1})^{-1} = acb^{-1}.$$ Thus, we must have $b^{-1}c^{-1} = cb^{-1}$ for all $b,c\in G$. In particular, for all $x\in G$, $x^{-2}= x^{-1}x^{-1} = xx^{-1} = 1$, so every element is of exponent $2$.

But if every element is of exponent $2$, then $x^{-1}=x$ for all $x$, so $\star$ is the same thing as $\cdot$. That is, you just get the same group you started with.

So the only time you get a group is if you get back the original group, which happens if and only if the square of every element of your group is trivial.