Symmetry group of a non abelian group manifold Reading an article

...$B$ is itself the manifold of some group $H$. It should be noted that, if $H$ is a non abelian group, the symmetry group $G$ of the group manifold is not $H$ but $H\times{}H$, since the group manifold can be transformed by either left or right multiplication.

I don't get the left or right argument
 A: First, a caveat: The "symmetry group of a manifold" is not, by itself, a well-defined concept.  Without some further structure on the manifold, the only groups that are completely determined by the manifold are infinite-dimensional ones such as its diffeomorphism group or homeomorphism group.  In order to identify a finite-dimensional "symmetry group," you have to say what you want the group action to preserve -- should it preserve a Riemannian or pseudo-Riemannian metric?  Or a Lie group structure? Or a symplectic structure? Or something else?
With that in mind, what seems to be going on here is this: If $H$ is any Lie group, then there's a natural action of $H\times H$ on itself by left and right multiplication:
$$(h_1,h_2)\centerdot h = (h_1)h(h_2)^{-1}.$$
But depending on the group, this action might or might not be effective (meaning that only the identity of $H\times H$ acts trivially on $H$).  In the special case in which $H$ is abelian, then every element of the diagonal subgroup $\Delta = \{(h,h): h\in H\}$ acts trivially, so in fact this is more properly considered as an action of the quotient group $(H\times H)/\Delta$.  This group is isomorphic to $H$ itself, and its action on $H$ is equivalent to the left action given by $h_1\centerdot h = h_1 h$.  This seems to be the sense in which the article is considering the symmetry group to be $H$ in the abelian case and $H\times H$ otherwise.  But this is not really correct either, because for some nonabelian groups there will still be elements of $H\times H$ that act trivially (the subgroup of $\Delta$ consisting of central elements of the group), so the "symmetry group" in general might be some quotient of $H\times H$.
In what sense is this "the symmetry group of $H$"?  There's not enough information to say. If $H$ is endowed with a bi-invariant metric, then the action of $H\times H$ (or whatever quotient of it acts effectively) preserves that metric. But not every Lie group has a bi-invariant metric.  On the other hand, every Lie group does have a left-invariant metric, but that will typically be preserved by $H\times \{\text{Id}\}$, not in general by all of $H\times H$. 
You could also ask about group actions that preserve the Lie group structure of $H$ -- the automorphism group of $H$.  But this isn't it.  It is true that the subgroup $\Delta$ acts on $H$ by automorphisms (conjugation), but that might not be the entire automorphism group. (The automorphisms in $\Delta$ are called inner automorphisms, and the other ones are outer.)
So the bottom line is that you have to read further in the article to figure out what "symmetry group" the author is talking about.
