## 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 $\langle \tau\sigma^3 \rangle$

• 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.

• nbloomf  On November 18, 2011 at 9:27 am

Thanks!