# 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 ?

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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 ${\rm SD}_{2n}$.

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A pair of non-isomorphic groups with equivalent ("the same") character tables and power maps is called a Brauer pair. -- Maybe this keyword could be mentioned in the answer. –  Stefan Kohl Oct 15 at 15:57
@StefanKohl feel free to edit the answer to mention Brauer pair. But $D_{2n}$ and $Q_{2n}$ do not have the same power maps, so presumably they do not form a Brauer pair? –  Derek Holt Oct 15 at 17:46
What description is it you say only three isomorphism classes satisfy? There are $4$ distinct $2$-groups of any given (large enough) order with a cyclic subgroup of index $2$ (your list is missing the quasidihedral group). –  Tobias Kildetoft Oct 15 at 18:04
@TobiasKildetoft: I specified that $|Z(G)|=2$. You can calculate $|Z(G)|$ from the character table. The quasidihedral group has a larger centre than the other three groups. (It also has a different number of conjugacy classes.) –  Derek Holt Oct 15 at 18:38
Ahh, thank you. I thought I had to have overlooked some extra condition. –  Tobias Kildetoft Oct 15 at 18:43
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