## Find the centralizer of each element in Dih(8), the quaternion group, Sym(3), and Dih(16)

Use the given subgroup lattices to find the centralizers of every element of the following groups.

1. $D_8$
2. $Q_8$
3. $S_3$
4. $D_{16}$

Lemma 1: Let $G$ be a group, let $x \in G$, and let $z \in Z(G)$. Then $C_G(xz) = C_G(x)$. Proof: Suppose $y \in C_G(x)$; then $yxz = xyz = xzy$, so $y \in C_G(xz)$. If $y \in C_G(xz)$, we have $yxz = xzy = xyz$, so that by cancellation, $yx = xy$ and we have $y \in C_G(x)$. $\square$

Lemma 2: Let $G$ be a group and $x,g \in G$. Then $C_G(gxg^{-1}) = g(C_G(x))g^{-1}$. Proof: Suppose $y \in C_G(x)$. Then $y = gzg^{-1}$ for some $z \in C_G(x)$. Then $(gzg^{-1})(gxg^{-1})$ $= gzxg^{-1}$ $= gxzg^{-1}$ $= (gxg^{-1})(gzg^{-1})$, so that $y \in C_G(gxg^{-1})$. Suppose $y \in C_G(gxg^{-1})$. Then $ygxg^{-1} = gxg^{-1}y$, so that $(g^{-1}yg)x = x(g^{-1}yg)$. Thus $g^{-1}yg \in C_G(x)$, and we have $y \in g C_G(x) g^{-1}$. $\square$

Lemma 3: Let $G$ be a group and let $g \in G$, $A \subseteq G$. Then $g \langle A \rangle g^{-1} = \langle g A g^{-1} \rangle$. Proof: Let $y \in g \langle A \rangle g^{-1}$. Then $y = gzg^{-1}$ where $z \in \langle A \rangle$; recall that we may write $z = a_1 a_2 \ldots a_n$ where for each $i$, either $a_i$ or $a_i^{-1}$ is in $A$. Then $z = a_1g^{-1}ga_2g^{-1}g\ldots g^{-1}ga_n$, so that $gzg^{-1} = (ga_1g^{-1})(ga_2g^{-1}\ldots(ga_ng^{-1})$. Hence $y \in \langle gAg^{-1} \rangle$. All of these steps are “if and only if”, so the sets are equal. $\square$

Lemma 4: Let $G$ be a group and let $a,b \in G$. If $\langle a \rangle = \langle b \rangle$, then $C_G(a) = C_G(b)$. Proof: Note that $b = a^k$ for some $k$, so that if $xa = ax$, we have $xb = bx$. Hence $C_G(a) \leq C_G(b)$. The other direction is similar. $\square$

1. $G = D_8$. Recall that $Z(D_8) = \{ 1, r^2 \}$.
 $x$ Reasoning $C_G(x)$ $1$ $1 \in Z(D_8)$ $D_8$ $r$ $\langle r \rangle \leq C_G(r)$, so $C_G(r)$ is either $\langle r \rangle$ or $D_8$. But $sr \neq rs$, so $s \notin C_G(r)$, hence $C_G(r) \neq D_8$. $\langle r \rangle$ $r^2$ $r^2 \in Z(D_8)$ $D_8$ $r^3$ $r^3 = r^{-1}$ $\langle r \rangle$ $s$ $\langle s \rangle \leq C_G(s)$ and $\langle r^2 \rangle \leq C_G(s)$ since $r^2 \in Z(G)$, so $C_G(s)$ is either $\langle s,r^2 \rangle$ or $D_8$. But $r \notin C_G(s)$ since $sr \neq rs$. $\langle s,r^2 \rangle$ $sr$ $\langle sr \rangle \leq C_G(sr)$ and $\langle r^2 \rangle \leq C_G(sr)$ since $r^2 \in Z(G)$, so $C_G(sr)$ is either $\langle sr, r^2 \rangle$ or $D_8$. But $rsr = s \neq sr^2$, so $r \notin C_G(sr)$. $\langle sr, r^2 \rangle$ $sr^2$ $sr^2 = s \cdot r^2$, so $C_G(sr^2) = C_G(s)$ by Lemma 1 above. $\langle s,r^2 \rangle$ $sr^3$ $sr^3 = sr \cdot r^2$, so $C_G(sr^3) = C_G(sr)$ by Lemma 1 above. $\langle sr,r^2 \rangle$
2. $G = Q_8$. Recall that $Z(Q_8) = \{ 1, -1 \}$.
 $x$ Reasoning $C_G(x)$ $1$ $1 \in Z(G)$ $Q_8$ $-1$ $-1 \in Z(G)$ $Q_8$ $i$ $\langle i \rangle \leq C_G(i)$ so $C_G(i)$ is either $\langle i \rangle$ or $Q_8$. But $j \notin C_G(i)$ since $ij = k \neq -k = ji$. $\langle i \rangle$ $-i$ $-i = i^{-1}$, so $C_G(-i) = C_G(i)$ $\langle i \rangle$ $j$ $\langle j \rangle \leq C_G(j)$ so $C_G(j)$ is either $\langle j \rangle$ or $Q_8$. But $k \notin C_G(j)$ since $jk = i \neq -i = kj$. $\langle j \rangle$ $-j$ $-j = j^{-1}$, so $C_G(-j) = C_G(j)$. $\langle j \rangle$ $k$ $\langle k \rangle \leq C_G(k)$, so $C_G(k)$ is either $\langle k \rangle$ or $Q_8$. But $i \notin C_G(k)$ since $ki = j \neq -j = ik$. $\langle k \rangle$ $-k$ $-k = k^{-1}$, so $C_G(-k) = C_G(k)$. $\langle k \rangle$
3. $S_3$
 $x$ $C_G(x)$ $1$ $S_6$ $(1\ 2)$ $\langle (1\ 2) \rangle \leq C_G((1\ 2))$, so $C_G((1\ 2))$ is either $S_6$ or $\langle (1\ 2) \rangle$. By a lemma to a previous exercise, there exists an element of $S_6$ which does not commute with $(1\ 2)$. $\langle (1\ 2) \rangle$ $(1\ 3)$ Note that $(1\ 3) = (2\ 3)(1\ 2)(2\ 3)$, so we can apply Lemmas 2 and 3. $\langle (1\ 3) \rangle$ $(2\ 3)$ Note that $(2\ 3) = (1\ 3)(1\ 2)(1\ 3)$, so we can apply Lemmas 2 and 3. $\langle (2\ 3) \rangle$ $(1\ 2\ 3)$ $\langle (1\ 2\ 3) \leq C_G((1\ 2\ 3))$, so $C_G((1\ 2\ 3))$ is either $\langle (1\ 2\ 3) \rangle$ or $S_6$. But $(1\ 2)$ does not commute with $(1\ 2\ 3)$. $\langle (1\ 2\ 3) \rangle$ $(1\ 3\ 2)$ $(1\ 3\ 2) = (1\ 2\ 3)^{-1}$ $\langle (1\ 2\ 3) \rangle$
4. $D_{16}$. Recall that, as we saw in a previous exercise, $r^4$ commutes with all elements of $D_{16}$.
 $x$ $C_G(x)$ $1$ $D_{16}$ $r$ $\langle r \rangle \leq D_{16}$, so $C_G(r)$ is either $D_{16}$ or $\langle r \rangle$. But $rs \neq sr$. $\langle r \rangle$ $r^2$ $\langle r^2 \rangle \leq C_G(r^2)$, so $C_G(r)$ is either $\langle r^2 \rangle$, $\langle s,r^2 \rangle$, $\langle r \rangle$, $\langle sr,r^2 \rangle$, or $D_{16}$. Note that $sr^2 \neq r^2 s$, and $rr^2 = r^2r$, and $srr^2 \neq r^2sr$. $\langle r \rangle$ $r^3$ $\langle r^3 \rangle = \langle r \rangle$, so Lemma 4 applies. $\langle r \rangle$ $r^4$ $r^4 \in Z(G)$. $D_{16}$ $r^5$ $\langle r^5 \rangle = \langle r \rangle$, so Lemma 4 applies. $\langle r \rangle$ $r^6$ $r^6 = r^2 r^4$ $\langle r \rangle$ $r^7$ $r^7 = r^{-1}$. $\langle r \rangle$ $s$ $\langle s \rangle \leq C_G(s)$, so $C_G(s)$ is either $\langle s \rangle$, $\langle s, r^4 \rangle$, $\langle s,r^2 \rangle$, or $D_{16}$. Now $\langle r^4 \rangle \leq C_G(s)$, and $r^2s \neq sr^2$. $\langle s,r^4 \rangle$ $sr$ $\langle sr \rangle \leq C_G(sr)$, so $C_G(sr)$ is either $\langle sr \rangle$, $\langle sr,r^4 \rangle$, $\langle sr,r^2 \rangle$, or $D_{16}$. Note that $r^4 sr = srr^4$ and $srr^2 \neq r^2sr$. $\langle sr,r^4 \rangle$ $sr^2$ $\langle sr^2 \rangle \leq C_G(sr^2)$, so $C_G(sr^2)$ is either $\langle sr^2 \rangle$, $\langle sr^2,r^4 \rangle$, $\langle s,r^2 \rangle$, or $D_{16}$. Note that $r^4sr^2 = sr^2r^4$ and $r^2sr^2 \neq sr^2r^2$. $\langle sr^2,r^4 \rangle$ $sr^3$ $\langle sr^3 \rangle \leq C_G(sr^3)$ and $\langle r^4 \rangle \leq C_G(sr^3)$. Note that $r^2 sr^3 \neq sr^3r^2$. $\langle sr^3,r^4 \rangle$ $sr^4$ Lemma 1 $\langle s,r^4 \rangle$ $sr^5$ Lemma 1 $\langle sr,r^4 \rangle$ $sr^6$ Lemma 1 $\langle sr^2,r^4 \rangle$ $sr^7$ Lemma 1 $\langle sr^3,r^4 \rangle$

• Bobby Brown  On September 19, 2010 at 8:07 pm

typo in centralizer of sr^3.

• nbloomf  On September 19, 2010 at 9:04 pm

Fixed. Thanks!

• Gobi Ree  On November 17, 2011 at 1:34 am

It seems that the typo in centralizer of sr^3 still exists.

• nbloomf  On November 17, 2011 at 10:55 am

Thanks!