Existence of a Lie subgroup Let $G=SU(k)\times T^1$, $S$ a subgroup of the center of $SU(k)$ ($Z(SU(k)\cong \mathbb{Z}_k$) and $\eta$ a homomorphism from $S$ into $T^1$. Suppose $(S, \eta)$ denotes the subgroup of $G$ contains elements of the form $(s , \eta(s))$ for all $s\in S$. We know that $SO(2k)$ has some subgroups in the form $G/(S,\eta)$ (for example $U(k)$). It seems (for the reasons not mentioned here) $SO(2k)\times T^1$ has no subgroup in the above form (nontrivial $S$) which is not contained in $SO(2k)$, is it true?
*$SO(2k)\cong SO(2k)\times \{e\}$
 A: Since you allow $\eta$ to be trivial, a counterexample is provided by $G = SU(4)$ with $S = Z(SU(4))$.
So, in other words, I'm claiming $SU(4)/S \times T^1$ is not isomorphic to a subgroup of $SO(8)$.  In fact, I'm claiming more:  $SU(4)/S$ is not isomorphic to a subgroup of $SO(8)$.  The proof will use some representation theory.
Recall the Dynkin Diagram of $SU(4)$ is $A_3$, consisting of 3 simple roots arranged in a line.  For a general Lie group, the irreducible representations are uniquely characterized by their highest weight vector with can be given in terms of natural numbers over each of the 3 simple roots.
Using http://www-math.univ-poitiers.fr/~maavl/LiE/form.html, computing the dimension of a module, we see that $SU(4)$ has exactly 3 irreducible representations of dimension less than or equal to $8$, given by putting $0$s over 2 of the simple roots and a 1 over the remaining simple root.
Putting a $1$ over either the first or last root gives a 4-d rep.  The first of these is the standard rep while the second is the first precomposed with complex conjugation.
Said another way, the first rep is $A\cdot v = Av$ while the second is $A\cdot v = \overline{A} v$.
Putting a $1$ over the middle simple root corresponds to the representation given by the usual double cover $SU(4)\rightarrow SO(6)$.
General representation theory tells us that a sum of representations if real iff its a sum of real irreducible plus pairs of other representations and their conjugates.  In particular, the only nontrivial homomorphisms from $SU(4)$ to $SO(8)$ are the standard homomorphism coming from interpreting $\mathbb{C}$ as $\mathbb{R}^2$ and the double cover $SU(4)\rightarrow SO(6)\subseteq SO(8)$.
Now, we simply need to check the kernels of both of these and make sure they are not all of $S$.  But, the first homomorphism has trivial kernel and the second has kernel generated by $-I$, so neither is all of $S$.  It follows that $SU(4)/S$ is not isomorphic to a subgroup of $SO(8)$.
(Incidentally, the adjoint action of $SU(4)$ on its algebra gives an embedding of $SU(4)/S$ into $SO(15)$.  I'm not sure if this is the smallest $SO(n)$ it embeds into or not.)
