The Lie algebra, $\frak{g}$, of a compact Lie group, $G$, is reductive, meaning its a direct sum of an abelian Lie algebra and a semisimple Lie algebra. This is not difficult to prove. Here's a proof
http://books.google.com/books?id=PAJmnm3DU1QC&pg=RA1-PA249#v=onepage&q&f=false
(The proof there uses an equivalent definition of reductive: every ideal has a complimentary ideal. But proving the equivalence is not difficult.)
If $\frak{g}$ is solvable, the semisimple part must be zero. So $\frak{g}$ is abelian. Therefore $G$ is a torus.
If I've skipping too many details, please let me know.
This is true: a compact connected Lie group with trivial center is a product of simple compact connected Lie groups with trivial center.
Suppose $G$ is a compact connected Lie group. Then $G$ has a cover of the form $G':=T^k\times G_1\times ... \times G_n$ where $T^k$ is a $k$-dimensional torus and the $G_i$ are simple simply connected Lie groups. In addition, there is a discrete subgroup $\Gamma\subseteq Z(G') = T^k\times Z(G_1)\times ....\times Z(G_n)$ for which $G'/\Gamma \cong G$. Here, the notation $Z(\cdot)$ refers to the center.
Now, suppose for a moment that $\Gamma$ is a proper subgroup of $Z(G')$. Select $g\in Z(G')\setminus \Gamma$.
Proposition: The element $g\Gamma\in G'/\Gamma$ is central. That is, $g\Gamma \in Z(G'/\Gamma).$
Proof: Select any $h\Gamma\in G'/\Gamma$. Then $$(g\Gamma)(h\Gamma) = (gh)\Gamma = (hg)\Gamma = (h\Gamma)(g\Gamma).$$ $\square$
So, if we assume additionally that $G$ is centerless, we deduce that $\Gamma = Z(G_1)\times ... \times Z(G_n)$ (and that $k=0$). But then, $$G'/\Gamma = (G_1\times ...\times G_n)/(Z(G_1)\times ....\times Z(G_n) = (G_1/Z(G_1))\times ...\times (G_n/Z(G_n)).$$
This shows that $G$ is a product of simple compact connected groups, but why are they centerless?
Proposition Suppose $H$ is a connected simple Lie group. Then $H/Z(H)$ is centerless.
Before proving this, note that the assumption that $H$ is connected is crucial. E.g., if $H = Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ is the Quaternion group, then $Z(Q_8) = \{\pm 1\}$, and $Q_8/Z(Q_8)\cong \mathbb{Z}_2\oplus \mathbb{Z}_2$, which has large center.
Proof: Let $hZ(H)\in Z(H/Z(H))$. Our goal is to show that this element must be the identity. That is, our goal is to show that $h \in Z(H)$. Choose any $gZ(H)\in H/Z(H)$. Then, by assumption, $hgZ(H) = ghZ(H)$, so $g^{-1}h^{-1}gh\in Z(H)$. Now, since $H$ is connected, there is a path $\gamma(t)$ in $H$ with $\gamma(0) = e$ (the identity) and $\gamma(1) = g$. Thus, we obtain a path $\alpha(t):=\gamma(t)^{-1} h^{-1}\gamma(t) h$ in $Z(H)$. Since $H$ is simple, $Z(H)$ is discrete, so $\alpha(t)$ is constant. Since $\alpha(0) = e$, we must therefore have $\alpha(1) = e$. That is, $g^{-1}h^{-1}gh = e$, so $gh = hg$. Thus, $h\in Z(H)$ as claimed. $\square$
Best Answer
No. Consider, for instance, the group $U(2)$. It is homeomorphic to $SU(2)\times U(1)$. However, $U(2)$ and $SU(2)\times U(1)$ are not isomorphic Lie groups.