Isomorphic Lie group structures on the same manifold Let $G$ be a smooth manifold. Suppose that we have two different smooth maps
$$
\mu_{1}:G\times G\rightarrow G,
$$
$$
\mu_{2}:G\times G\rightarrow G,
$$
that give $G$ a Lie group structure. I was wondering if $(G,\mu_{1})$ and $(G,\mu_{2})$ could be isomorphic as Lie groups.
 A: In general no. For example, take
$$G=\mathrm{SU}(2)\times\mathbb{R}^3.$$
One Lie group structure comes from the usual Lie group structure on $\mathrm{SU}(2)$ together with addition on $\mathbb{R}^3$. Another one comes from a diffeomorphism with $\mathrm{SL}(2,\mathbb{C})$. Namely, identifying $\mathbb{R}^3$ with $\mathfrak{su}(2)$ we have
$$G=\mathrm{SU}(2)\times\mathfrak{su}(2)\overset{\cong}{\longrightarrow}\mathrm{SL}(2,\mathbb{C}),\quad(g,X)\longmapsto ge^{iX}.$$
(This is the Polar Decomposition.) The usual Lie group structure on $\mathrm{SL}(2,\mathbb{C})$ gives a different, non-isomorphic, Lie group structure on $G$.
To see that they are non-isomorphic, note that in the first case, the center is $\mathbb{Z}_2\times\mathbb{R}^3$ while in the second case, the center is $\mathbb{Z}_2$.
A: Every solvable $n$-dimensional connected Lie group is diffeomorphic to $\mathbb{R}^n$, but for $n=2$, there is the commutative Lie group $\mathbb{R}^2$ and the connected component of the group of affine transformations of the line which are not isomorphic.
A: Did you really mean to ask if they "could be isomorphic"? The answer to that question is certainly yes. For example, let $G=\mathbb R$, let $\mu_1$ be the usual addition, and let $\mu_2$ be $\mu_2(x,y) = x+y+1$. Then $\mu_2$ defines a Lie group structure (with $-1$ as the zero element), and it's isomorphic to the usual one by the map $\phi\colon \mathbb R\to \mathbb R$ given by $\phi(x)=x+1$. You can check that $\phi(\mu_2(x,y)) = \mu_1(\phi(x),\phi(y))$. 
But if you meant to ask whether they must be isomorphic, then the answers by Tsemo Aristide and Spenser show clearly that the answer is no.
