## Use the subgroup lattice to compute normalizers in the quasidihedral group of order 16

Use the subgroup lattice of $G = QD_{16}$ to help find the normalizers (1) $N_G(\langle \tau\sigma \rangle)$ and (2) $N_G(\langle \tau,\sigma^4 \rangle)$.

We computed the subgroup lattice of $QD_{16}$ in this previous exercise.

1. (Note that $\tau\sigma = \sigma^3\tau$.) We have $\langle \tau\sigma \rangle \leq N_G(\langle \tau\sigma \rangle)$, so that this normalizer is either $\langle \tau\sigma \rangle$, $\langle \sigma^2, \tau\sigma \rangle$, or $QD_{16}$.

Note that $\sigma^{-2}(\tau\sigma)\sigma^2 = \sigma^7\tau = \tau\sigma^5$ $= (\tau\sigma)^3$, so that $\langle \sigma^2, \tau\sigma \rangle \leq N_G(\langle \tau\sigma \rangle)$, but $\tau^{-1}(\tau\sigma)\tau = \tau\sigma^3 \notin \langle \tau\sigma \rangle$. So $N_G(\langle \tau\sigma \rangle) = \langle \sigma^2, \tau\sigma \rangle$.

2. We have $\langle \tau, \sigma^4 \rangle \leq N_G(\langle \tau, \sigma^4 \rangle)$, so that this normalizer is either $\langle \tau, \sigma^4 \rangle$, $\langle \tau, \sigma^2 \rangle$, or $QD_{16}$.

Now $\langle \tau, \sigma^4 \rangle = \{ 1, \tau, \sigma^4, \tau\sigma^4 \}$. Note that $\sigma^2$ is in this normalizer since $\sigma^2 \tau \sigma^{-2} = \tau \sigma^4$, $\sigma^2 \sigma^4 \sigma^{-2} = \sigma^4$, and $\sigma^2 \tau \sigma^4 \sigma^{-2} = \tau$. However, $\sigma$ is not in this normalizer since $\sigma \tau \sigma^{-1} = \tau \sigma^2$. So $N_G(\langle \tau,\sigma^4 \rangle) = \langle \tau,\sigma^2 \rangle$.

I think you confused the normalizer with the centralizer in problem 1. Since $\sigma^2 \tau \sigma \sigma^{-2}=\tau \sigma^5 \in \langle \tau \sigma \rangle$ and $\sigma \tau \sigma \sigma^{-1}=\tau \sigma^3 \notin \langle \tau \sigma \rangle$, the normalizer should be $\langle \sigma^2, \tau \sigma \rangle$.