a question about the semidihedral group? My question is simple:
If a group $G$ has the same character table with the semidihedral group $SD_{2n}$, are $G$ and $SD_{2n}$ isomorphic ?
 A: I believe the answer is yes. When $n=2^k$, the group $G = {\rm SD}_{2n}$ is defined by the presentation $\langle x,y \mid y^n=x^2=1, y^{-1}xy=x^{m-1} \rangle$, where $m = n/2=2^{k-1}$.
Since $G$ has an abelian subgroup of index 2, its irreducible characters all have degree 1 or 2. Consider a faithful irreducible representation $\rho$ of degree 2. Since $x$ is conjugate to $x^{m-1}$, the eigenvalues of $\rho(x)$ must be $w$ and $w^{m-1} = -w^{-1}$, where $w$ is a primitive $n$-th root of 1. So its character value on $x$ is $w-w^{-1}$, which is purely imaginary.
Since $n \ge 8$, it is not hard to see that this is the only way we can express $w-w^{-1}$ as the sum of two roots of 1, and so any group $H$ having the same character table as $G$ must have an element of order $n$. Moreover, $M$ is nonabelian with a cyclic subgroup of order $n$ and has centre of order 2.
$2$-groups with a cyclic subgroup of index 2 have been classified, and there are only three isomorphism types that fit that description, $G = {\rm SD}_{2n}$, the dihedral group $D_{2n}$ and the generalized quaternion group $Q_{2n}$.
But $D_{2n}$ and $Q_{2n}$ do have the same character tables, and all of their characters are real (their elements of order $n$ have character values $w+w^{-1}$ rather than $w-w^{-1}$), so they are not the same as that of ${\rm SD}_{2n}$.
