Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.5
For each of the following subgroups $A$ of a given group $G$, show that $C_G(A) = A$ and $N_G(A) = G$.
(1) $G = S_6, A = \{ 1, (1\ 2\ 3), (1\ 3\ 2) \}$
(2) $G = D_8, A = \{ 1, s, r^2, sr^2 \}$
(3) $G = D_{10}, A = \{ 1, r, r^2, r^3, r^4 \}$
Solution: First we prove some lemmas.
Lemma 1: Let $G$ be a group and $A \subseteq G$. Then $C_G(A) \subseteq N_G(A)$.
Proof: Suppose $x \in C_G(A)$. Then $xax^{-1} = a$ for all $a \in A$; in particular, $xAx^{-1} = A$. Thus $x \in N_G(A)$. $\square$
Lemma 2: Let $G$ be a group and $A \subseteq G$. Then $C_G(A) = \bigcap_{x \in A} C_G(x)$.
Proof: ($\subseteq$) If $y \in C_G(A)$, we have $yxy^{-1} = x$ for all $x \in A$. Thus $y \in C_G(x)$ for all $x \in A$, hence $y \in \bigcap_{x \in A} C_G(x)$. ($\supseteq$) If $y \in \bigcap_{x \in A} C_G(x)$, we have $yxy^{-1} = x$ for all $x \in A$. Thus $y \in C_G(A)$. $\square$
(1) By Exercise 2.2.4 and Lemma 2, we have that $C_G(A) = A$ by inspection. Now by Lemma 1, $A \subseteq N_G(A)$. Moreover, $$(1\ 2)(1\ 2\ 3)(1\ 2) = (1\ 3\ 2)$$ and $$(1\ 2)(1\ 3\ 2)(1\ 2) = (1\ 2\ 3),$$ so that $(1\ 2) \in N_G(A)$. Thus $4 \leq |N_G(A)| \leq 6$, and by Lagrange’s Theorem, $|N_G(A)|$ divides 6. Thus $N_G(A) = S_3$.
(2) By Exercise 2.2.4 and Lemma 2, we have that $C_G(A) = A$ by inspection. Now by Lemma 1, $A \subseteq N_G(A)$. Moreover, $rsr^3 = sr^2$, $rr^2r^3 = r^2$, and $rsr^2r^3 = s$, so that $r \in N_G(A)$. So $5 \leq |N_G(A)| \leq 8$, and by Lagrange’s Theorem we have $N_G(A) = D_8$.
(3) In this case $A$ is cyclic, so that $A \subseteq C_G(A)$. So $5 \leq |C_G(A)|$. But note that $srs = r^4$, so that $s \notin C_G(A)$. By Lagrange’s Theorem the order of $C_G(A)$ must divide the order of $D_{10}$, but we see that $5 \leq |C_G(A)| \leq 9$. Thus $C_G(A) = A$. Now $A \subseteq N_G(A)$ by Lemma 1. Moreover, $sr^2s = r^3$, $sr^3s = r^2$, and $sr^4s = r$, so that $s \in N_G(A)$. Again by Lagrange, $N_G(A) = D_{10}$.
