Exercise 2.2.5 (Comparison of $C_G(A)$ and $N_G(A)$ in examples)

Proof.

Note: if we know that a subgroup with $|G|∕2$ elements is normal in $G$, then the subgroup $A$ is normal in $G$ in cases (a), (b) and (c). Therefore $N_G(A)=G$. We give below the corresponding elementary verifications of this fact.

(a)

$G=S_3$ and $A=\{1,(123),(132)\}=\{1,\sigma ,{\sigma }^2\}=⟨\sigma ⟩$, where $\sigma =(123)$

By Exercise 4, $C_G(1)=G,C_G(\sigma )=A,C_G({\sigma }^2)=A$. Therefore

$$C_G(A)=\underset{a\in A}{\bigcap }{C}_G(a)=G\cap A\cap A=A.$$

We know that $C_G(A)\le N_G(A)$. Moreover, since $\tau (abc){\tau }^{-1}=(\tau (a)\tau (b)\tau (c))$, we obtain for $\tau =(12)$

$$\begin{array}{llll}\hfill \mathit{\tau A}{\tau }^{-1}& =\{1,(\tau (1)\tau (2)\tau (3)),(\tau (1)\tau (3)\tau (2)\}& \hfill & \\ \hfill & =\{1,(213),(231)\}& \hfill & \\ \hfill & =\{1,(132),(123)\}& \hfill & \\ \hfill & =A.& \hfill & \end{array}$$

Therefore $\tau \in N_G(A)$, so $|N_G(A)|\ge 4$, where $|N_G(A)|$ divides $6$ by Lagrange’s Theorem. This gives

$$G=N_G(A).$$

(b)

$G=D_8$ and $A=\{1,s,r^2,s{r}^2\}=⟨s,r^2⟩$.

(We know by Exercise 4, or Exercise 2.1.3, that $A$ is a subgroup of $D_8$.)

By Exercise 4, since $s{r}^2=r^{2}s$, $C_G(1)=G,C_G(s)=A,C_G(r^2)=G,C_G(s{r}^2=A)$, therefore

$$C_G(A)=\underset{a\in A}{\bigcap }{C}_G(a)=G\cap A\cap G\cap A=A.$$

As in part (a), $C_G(A)\subseteq N_G(A)$. Moreover

$$\begin{array}{llll}\hfill \mathit{rA}{r}^{-1}& =\{1,\mathit{rs}{r}^{-1},r(r^2)r^{-1},r(s{r}^2)r^{-1}\}& \hfill & \\ \hfill & =\{1,s{r}^3{r}^{-1},r^2,s{r}^3{r}^2{r}^{-1}\}& \hfill & \\ \hfill & =\{1,s{r}^2,r^2,s\}& \hfill & \\ \hfill & =A.& \hfill & \end{array}$$

Therefore $r\in N_G(A)$. So $N_G(A)$ has at least $5$ elements. By Lagrange’s theorem, $A=N_G(A)$.

(c)

$G=D_{10}$ and $A=\{1,r,r^2,r^3,r^4\}=⟨r⟩$. Since $A=⟨r⟩$, $A\le C_G(A)$. By Lagrange’s Theorem, $5$ divides $|C_G(A)|$ which divides $10$, so $|C_G(A)|=5$ or $10$.

If $|C_G(A)|=10$, then $C_G(A)=D_{10}$, but $\mathit{sr}{s}^{-1}=r^4$, so $s\notin C_G(A)$. This is impossible, so $|C_G(A)|=4$. This shows that

$$C_G(A)=A.$$

As above, $C_G(A)\le N_G(A)$, so $|N_G(A)|\ge 4$. Moreover

$$\begin{array}{llll}\hfill \mathit{sA}{s}^{-1}& =⟨\mathit{sr}{s}^{-1}⟩& \hfill & \\ \hfill & =⟨r^{-1}⟩& \hfill & \\ \hfill & =A.& \hfill & \end{array}$$

Therefore $s\in N_G(A)$, hence $|N_G(A)|\ge 5$ and

$$N_G(A)=G.$$