Compute the centralizer of every element in the quasidihedral group of order 16

Use the subgroup lattice of QD_{16} to find the centralizer of every element of QD_{16}.


Recall that QD_{16} = \langle \sigma,\tau \ |\ \tau^2 = \sigma^8 = 1, \tau\sigma = \sigma\tau^3 \rangle. We computed the subgroup lattice of G = QD_{16} in a previous exercise.

Note that we can use the relations ]tau^2 = \sigma^8 = 1 to write any word in \sigma, \tau in QD_{16} using only positive powers. Then we can use the relation \sigma\tau = \tau\sigma^3 to express every element of G in the form \tau^a\sigma^b for some a and b, and the first two relations give us that 0 \leq a < 2 and 0 \leq b < 8. Since we know that the order of this group is 16, all such elements are distinct.

Lemma 1: Let G be a group, x \in G, and A \subseteq G. If ax = xa for all a \in A, then x \in C_G(\langle A \rangle). Proof: Every element of \langle A \rangle is a word in A; by hypothesis, then, x commutes with every element of \langle A \rangle. \square

Lemma 2: \sigma^4 \in Z(QD_{16}). Proof: We have \sigma^4 \sigma = \sigma \sigma^4 and \sigma^4 \tau = \tau \sigma^{20} = \tau \sigma^4. By Lemma 1, then, \sigma^4 \in C_G(\langle \tau,\sigma \rangle) = Z(G). \square

We will make judicious use of the lemmas of a previous exercise concerning centralizers.

x C_G(x)
1 1 \in Z(G) QD_{16}
\sigma We have \langle \sigma \rangle \leq C_G(\sigma), so C_G(\sigma) is either \langle \sigma \rangle or QD_{16}. But \tau\sigma \neq \sigma\tau so \tau \notin C_G(\sigma). \langle \sigma \rangle
\sigma^2 \langle \sigma^2 \rangle \leq C_G(\sigma^2), so that C_G(\langle \sigma^2 \rangle is either \sigma^2 \rangle, \langle \sigma^2,\tau \rangle, \langle \sigma \rangle, \langle \sigma^2,\sigma\tau \rangle, or QD_{16}. Now \sigma \in C_G(\sigma^2) and \tau \notin C_G(\sigma^2). \langle \sigma \rangle
\sigma^3 Lemma 4 to a previous exercise. \langle \sigma \rangle
\sigma^4 \sigma^4 \in Z(G) QD_{16}
\sigma^5 Lemma 1 to a previous exercise \langle \sigma \rangle
\sigma^6 Lemma 1 to a previous exercise \langle \sigma \rangle
\sigma^7 Lemma 4 to a previous exercise \langle \sigma \rangle
\tau We have \langle \tau, \sigma^4 \rangle \leq C_G(\tau), but \sigma^2 \notin C_G(\tau). \langle \tau,\sigma^4 \rangle
\tau\sigma \sigma^2 \notin C_G(\tau\sigma) \langle \tau\sigma \rangle
\tau\sigma^2 \langle \sigma^2\tau,\sigma^4 \rangle \leq C_G(\tau\sigma^2), and \tau \notin C_G(\tau\sigma^2). \langle \tau\sigma^2,\sigma^4 \rangle
\tau\sigma^3 Lemmas 2 and 3 to a previous exercise. \langle \tau\sigma^3 \rangle
\tau\sigma^4 Lemma 1 to a previous exercise \langle \tau,\sigma^4 \rangle
\tau\sigma^5 Lemma 4 to a previous exercise \langle \tau\sigma \rangle
\tau\sigma^6 \langle \tau\sigma^6,\sigma^4 \rangle \leq C_G(\tau\sigma^6), but \tau \notin C_G(\tau\sigma^6) \langle \tau\sigma^6,\sigma^4 \rangle
\tau\sigma^7 Lemma 4 to a previous exercise \langle \tau\sigma^3 \rangle
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Comments

  • Gobi Ree  On November 17, 2011 at 9:12 pm

    typo: ]tau — \tau,
    \tau^2\sigma^2 in the tau’s row of the table.

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