# Can nonabelian groups still have isomorphic left- and right- multiplication?

A group $G$ is is abelian if

$$ab = ba$$

for all pairs of elements $a,b \in G$, i.e., if multiplication commutes. Another way of stating this condition is that the identity map $\mathrm{id}$ gives an isomorphism between left- and right- multiplication. In other words, we can think of left-and right- multiplication as group actions

$$L: G \rightarrow \mathrm{Sym}(G);\ L(a)\cdot b := ab,$$

$$R: G \rightarrow \mathrm{Sym}(G);\ R(a)\cdot b := ba.$$

In an abelian group we have $L(a)\cdot b = ab = ba = R(a)\cdot b$. Hence,

$$L(\mathrm{id}(a))=R(a).$$

One might consider a weaker condition: there is some isomorphism $\varphi:G \rightarrow G$ (not necessarily the identity) such that

$$L(\varphi(a))=R(a).$$

for all $a \in G$. (In other words, if we really didn't like multiplying on the right, we could always multiply on the left by $\varphi(a)$ instead without losing anything.)

Is there a standard name for this weaker condition? When can it be satisfied (if ever)? For some standard examples (e.g., the quaternions) I can work out that no such isomorphism exists, but I don't know what to say about the general case.

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@user1612: Left (or right) multiplication are not automorphisms of $G$! They do not send the identity to the identity. They are permutations of $G$; the map does not go from $G$ to $\mathrm{Aut}(G)$, but to $S_G$. – Arturo Magidin Nov 2 '10 at 3:59
Right! Good catch -- thanks. – user1612 Nov 2 '10 at 4:05
@user1612: And also note that while $L$ is a group homomorphism, $R$ is not; $R$ is a group anti-homomorphism. – Arturo Magidin Nov 2 '10 at 4:21

In particular, you would have $L(\varphi(a))\cdot e = R(a)\cdot e$, that is, $\varphi(a)=a$. So $\varphi=\mathrm{id}$.
Note that $L$ is a one-to-one homomorphism from $G$ to $S_G$, while $R$ is a one-to-one homomorphism from $G^{\mathrm{op}}$ to $S_G$ ($R(ab) = R(b)\circ R(a)$, so you have an anti-homomorphism if you work with $G$).
You could ask more generally if these two subgroups of $S_G$ are conjugate. That is, do there exist permutations $\pi$ and $\psi$ of $G$ such that for every $g\in G$, $\pi^{-1}\circ L(g) \pi = R(\psi(g))$ ? In fact, they always are: take $\pi=\psi$ to be the map that sends every element $x$ to $x^{-1}$ (which is, after all, an isomorphism from $G$ to $G^{op}$!): $\pi^{-1}(L(g)(\pi(x))) = \pi^{-1}(L(g)(x^{-1})) = \pi^{-1}(gx^{-1}) = xg^{-1}) = R(g^{-1})(x).$ So $L(G)$ and $R(G)$ are always conjugate in $S_G$, whether $G$ is abelian or not.