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.

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Comments

  • Gobi Ree  On November 17, 2011 at 9:23 pm

    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.

    • nbloomf  On November 18, 2011 at 12:09 pm

      You’re right. Thanks!

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