## Compute the normalizer of each subgroup in Sym(3) and the quaternion group

In each of the following groups find the normalizer of each subgroup.

(1) $S_3$ (2) $Q_8$

Lemma 1: For all groups $G$, $N_G(1) = G$. Proof: If $g \in G$, we have $g \cdot 1 \cdot g^{-1} = 1$, so $G \subseteq N_G(1)$. $\square$

Lemma 2: For all groups $G$, $N_G(G) = G$. Proof: Clearly if $g \in G$ we have $gGg^{-1} = G$. $\square$

Lemma 3: Let $G$ be a group and $A \subseteq Z(G)$. Then $N_G(A) = G$. Proof: Let $g \in G$ and $a \in A$. Then $gag^{-1} = a$, so that $gAg^{-1} = A$. Hence $G \subseteq N_G(A)$. $\square$

Lemma 4: Let $G$ be a group, $h \in G$, and $A \subseteq G$. Then $N_G(hAh^{-1}) = h N_G(A) h^{-1}$. Proof: We have $g \in N_G(hAh^{-1})$ iff $ghAh^{-1}g^{-1} = hAh^{-1}$, iff $(h^{-1}gh)A(h^{-1}gh) = A$, iff $h^{-1}gh \in N_G(A)$, iff $g \in h N_G(A) h^{-1}$. $\square$

Lemma 5: If $H \leq G$ is a subgroup, then $H \leq N_G(H)$. Proof: Clearly if $h \in H$, $hHh^{-1} = H$. $\square$

Lemma 6: Let $G$ be a group, $a \in G$, and $g \in G$. If $\langle gag^{-1} \rangle = \langle a \rangle$, then $g \in N_G(\langle a \rangle)$. Proof: We saw in a lemma to a previous exercise that $g \langle a \rangle g^{-1} = \langle gag^{-1} \rangle = \langle a \rangle$, so that $g \in N_G(\langle a \rangle)$. $\square$

1. $G = S_6$
 $H$ Reasoning $N_G(H)$ $1$ Lemma 1 $S_3$ $\langle (1\ 2) \rangle$ $\langle (1\ 2) \rangle \leq N_G(\langle (1\ 2) \rangle)$, so $N_G(\langle (1\ 2) \rangle)$ is either $\langle (1\ 2) \rangle$ or $S_3$. But $(1\ 2\ 3)(1\ 2)(1\ 3\ 2) = (2\ 3)$. $\langle (1\ 2) \rangle$ $\langle (1\ 3) \rangle$ $(1\ 3\ 2)(1\ 2)(1\ 2\ 3) = (1\ 3)$, so Lemma 4 applies. $\langle (1\ 3) \rangle$ $\langle (2\ 3) \rangle$ $(1\ 2\ 3)(1\ 2)(1\ 3\ 2) = (2\ 3)$, so Lemma 4 applies. $\langle (2\ 3) \rangle$ $\langle (1\ 2\ 3) \rangle$ $\langle (1\ 2\ 3) \rangle \leq N_G(\langle (1\ 2\ 3) \rangle)$, so $N_G(\langle (1\ 2\ 3) \rangle)$ is either $\langle (1\ 2\ 3) \rangle$ or $S_3$. Note that $(1\ 2)(1\ 2\ 3)(1\ 2) = (1\ 3\ 2)$ and $(1\ 2)(1\ 3\ 2)(1\ 2) = (1\ 2\ 3)$, so that $(1\ 2) \in N_G(\langle (1\ 2\ 3) \rangle)$. Lagrange’s Theorem applies. $S_3$ $S_3$ Lemma 2 $S_3$
2. $G = Q_8$
 $H$ Reasoning $N_G(H)$ $1$ Lemma 1 $Q_8$ $\langle -1 \rangle$ Lemma 3 $Q_8$ $\langle i \rangle$ We have $\langle i \rangle \leq N_G(\langle i \rangle)$, so $N_G(\langle i \rangle)$ is either $\langle i \rangle$ or $Q_8$. Note that $jij^{-1} = i^{-1}$, so Lemma 6 applies, as does Lagrange’s Theorem. $Q_8$ $\langle j \rangle$ We have $\langle j \rangle \leq N_G(\langle j \rangle)$, so $N_G(\langle j \rangle)$ is either $\langle j \rangle$ or $Q_8$. Note that $kjk^{-1} = j^{-1}$, so Lemma 6 applies, as does Lagrange’s Theorem. $Q_8$ $\langle k \rangle$ We have $\langle k \rangle \leq N_G(\langle k \rangle)$, so $N_G(\langle k \rangle)$ is either $\langle k \rangle$ or $Q_8$. Note that $iki^{-1} = k^{-1}$, so Lemma 6 applies, as does Lagrange’s Theorem. $Q_8$ $Q_8$ Lemma 2 $Q_8$

• Gobi Ree  On November 17, 2011 at 1:56 am

Typo: the normalizer of $\langle (1\ 2\ 3) \rangle$ will be S_3.

• nbloomf  On November 17, 2011 at 10:49 am

Thanks!

• Gobi Ree  On November 17, 2011 at 2:03 am

WordPress deleted my symbols. I meant that $latex$.

• Gobi Ree  On November 17, 2011 at 2:04 am

Omg, I don’t know how to write the subgroup generated by (123).

• nbloomf  On November 17, 2011 at 10:51 am

WordPress automagically interprets less-than and greater-than symbols as opening and closing an HTML tag; to subvert this, type &lt; and &gt;.