Is this a feasible way to show that a group $G$ is isomorphic to its opposite group $G^{op}$?

Question

Let $G$ be a group with group operation $*$. Let $G^{op}$ be a group that shares the same base set with $G$ but the group operation $*^{'}$ is such that $a*^{'}b = b*a$. We would like to show that $G$ is isomorphic to $G^{op}$.

The standard way is to construct mapping $f: G \to G^{op}$ by $f(x)=x^{-1}$. This constitutes the isomorphism.

But I had an alternative idea which I am not sure whether it works.

Construct mapping $\varphi : G \to G^{op}$. (We assume $G$ is non-trivial of course.)Start with $\varphi(1_G)= 1 _{G^{op}}$. Choose a non-identity $g_1 \in G$, let $\varphi(g_1)=g_1$, and in general $\varphi(g_1 ^ k) = g_1 ^k, \forall k \in \mathbb Z$. Let the cyclic group generated by $g_1$ be named $G_1$.

Now we take an element $g_2 \notin G_1$. Let $\varphi(g_2^k)=g_2^k$. If an element of $G$ can be written as $g_1 ^ {k_1} g_2 ^{k_2}g_1^{k_3}....g_{1 \mbox{ or }2}^{k_\alpha}$, we map it to $ g_{1 \mbox{ or }2}^{k_\alpha}....g_1^{k_3}g_2 ^{k_2}g_1 ^ {k_1}$. This constructs the mapping on the subgroup $G_2$ generated by $g_1$ and $g_2$.

In general, if we have constructed the group $G_\gamma$ generated by a set of generators index by set $\Gamma$ $g_1, ... g_\gamma$, we introduce an element $g_{\gamma +1}$ such that $g_{\gamma+1} \notin G_{\gamma}$, we map the element $g_{i_1}^{k_1}g_{i_2}^{k_2} .... g_{i_\alpha}^{k_\alpha}$ to $g_{i_\alpha}^{k_\alpha} ... g_{i_2}^{k_2}g_{i_1}^{k_1}$ (note that $\alpha$ is finite). We do some sort of transfinite induction (I suppose?) to finish this off, and every element in $G$ can be represented as a word from a finite alphabet drawn from the set of generators. Then I think we should have constructed an anti-automorphism. Does this work?

Thank you.

Wilson

Answer 1

As Heinirich pointed out, the element of $G$ many not have unique expression in terms of generators; on the other hand consider $S_4$.

$$(12)(123)=(23) \mbox{ and also } (42)(423)=(23).$$ So while defining $f((23))$, on one hand we will have $$f(23)=(123)(12)=(13...)(4)$$ but on the other hand $$f(23)=(423)(42)=(43..)(1),$$ so $f$ no longer remains well-defined.

Answer 2

Take an isomorphism of $G$ onto $G^{op}$. Compose it with the isomorphism $x\mapsto x^{-1}$. You get an automorphism of $G$. All isomorphisms of $G$ onto $G^{op}$ are obtained by the reverse process.

Apply this remark to your candidate. You get an automorphism $\sigma$ of $G$ which sends a set of generators to their inverses. In particular, you must have $\sigma^2=1$. Also, $\sigma$ must induce $-1$ on the maximal abelian quotient of $G$, hence act by $(-1)^n$ on the $n$-th term of the graded Lie algebra associated to the lower central series. This obviously puts strong constraints on $G$: for example, if it is nonabelian and you take its quotient by $[G,[G,G]]$, you get a central extension which is split away from $2$.

On the other hand, there are obviously examples beyond abelian groups, e.g. free groups (or quotients of such by any characteristic subgroup).

It's interesting to know whether there are examples within finite simple groups.