Structure of semi direct product.

Question

I want to verify that structure of group $Q_8 \rtimes C_2$, i.e. semi direct product of Quaternion group of order $8$ and $C_2 = \{1, a\}$ (cyclic group of order $2$) can be defined like: If we write $$Q_8 = \{\pm 1, \ \pm x,\ \pm y,\ \pm xy\}$$ Define $$h:C_2 \to Aut(Q_8): a \mapsto g(a) = \begin{cases}x \to y\\ y \to x\\ -1\to -1\end{cases} $$ The define product in above group as $$(d_1, b_1)*(d_2, b_2) = (d_1d_2, b_1(g(d_1)(b_2))$$ Is it the correct representation or am I missing something?

Answer 1

Your group is a nontrivial semidirect product $Q_8\rtimes C_2$ isomorphic to the semidihedral group $$SD_{16} = \langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3\rangle$$ There is an isomorphism sending $a \mapsto (x,z)$ and $b\mapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:

• $(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.
• $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
• $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$

If you instead wanted $Q_{16} = Q_8\rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.