$GL(n,\mathbb R)$ not isomorphic to $SL(n,\mathbb R)\times \mathbb R^\ast$ when $n>1$

Question

If $GL(n,\mathbb R) \cong SL(n,\mathbb R) \times \mathbb R^\ast$, then the centers are isomorphic. When $n$ is even, this means that \begin{align*} \mathbb R^\ast &\cong \{\lambda I_n \mid \lambda\in\mathbb R^\ast\} \\ &= Z(GL(n,\mathbb R)) \\ &\cong Z(SL(n,\mathbb R)\times\mathbb R^\ast) \\ &= Z(SL(n,\mathbb R)) \times Z(\mathbb R^\ast) \\ &= \{\pm I_n\}\times\mathbb R^\ast. \end{align*} But this argument doesn't apply when $n>1$ is odd. For this case, I tried to use that the claim fails for $n-1$, but I haven't had success. Why does it fail for $n$ odd?

Answer 1

The claim doesn't fail for $n$ odd. If we identify $\Bbb R^\times$ with the invertible scalar matrices in ${\rm GL}_n(\Bbb R)$ then we easily check that ${\rm GL}_n(\Bbb R)={\rm SL}_n(\Bbb R)\Bbb R^\times$, that $[{\rm SL}_n(\Bbb R),\Bbb R^\times]=1$ and ${\rm SL}_n(\Bbb R)\cap\Bbb R^\times=1$, so we conclude that ${\rm GL}_n(\Bbb R)$ is an internal direct product of ${\rm SL}_n(\Bbb R)$ and $\Bbb R^\times$. The first and third facts both fail if the $n$ is even, so this doesn't work then.

Wikipedia even lists it in the example section of its article on direct products of groups.