Exercise 2.5.8 (Normalizer of each subgroup of $S_3$, of $Q_8$)

Question

In each of the following groups find the normalizer of each subgroup:
$$	ext{\bf(a)}\ S_3 \qquad	ext{\bf(b)}\ Q_8.$$

Richard Ganaye · Accepted Answer

\begin{proof} Normalizer of each subgroup of $G$.
\begin{enumerate}
\item[{\bf(a)}] $G = S_3$.

The normalizer of $\langle (1\ 2)angle$ in $G$ contains $(1\ 2)$, but not $(1\ 2 \ 3)$, so
$$\langle (1\ 2) angle  \leq N_G(\langle (1\ 2)angle) < G.$$
Using the lattice of subgroups of $S_3$, we obtain
$$N_G(\langle (1\ 2)angle) = \langle (1\ 2)angle.$$
The normalizer of $\langle (1\ 2\ 3)angle)$ in $G$ contains $(1\ 2\ 3)$ and $(1\ 2)$, therefore
$$N_G(\langle (1\ 2\ 3)angle) = S_3.$$
More generally,
\begin{center}
\begin{tabular}{|c|ccccc|}
\hline
 $H$  & $\langle() angle$ & $\langle (1\ 2 )angle$ & $\langle (1\ 3 )angle$  & $\langle (2\ 3 )angle$ & $\langle (1\ 2 \ 3)angle$  \
\hline
 $N_G(H)$&$S_3$  &$\langle  1, 2angle$  & $\langle (1\ 3 )angle$  & $\langle (2\ 3 )angle$ & $S_3$  \
\hline
\end{tabular}
\end{center}

\item[{\bf(b)}] $G = Q_8$.
\begin{center}
\begin{tabular}{|c|ccccc|}
\hline
 $H$  & $\langle 1 angle$ & $\langle -1 angle$  & $\langle i angle$ &$\langle j angle$& $\langle k angle$  \
\hline
 $N_G(H)$&$Q_8$  &$Q_8$  & $Q_8$  & $Q_8$ & $Q_8$\
 \hline
 \end{tabular}
\end{center}
All subgroups of $Q_8$ are normal!
\end{enumerate}

\end{proof}


$H$	$⟨()⟩$	$⟨ (1 2) ⟩$	$⟨ (1 3) ⟩$	$⟨ (2 3) ⟩$	$⟨ (1 2 3) ⟩$

$N_{G} (H)$	$S_{3}$	$⟨ 1, 2 ⟩$	$⟨ (1 3) ⟩$	$⟨ (2 3) ⟩$	$S_{3}$


$H$	$⟨ 1 ⟩$	$⟨ - 1 ⟩$	$⟨ i ⟩$	$⟨ j ⟩$	$⟨ k ⟩$

$N_{G} (H)$	$Q_{8}$	$Q_{8}$	$Q_{8}$	$Q_{8}$	$Q_{8}$

Exercise 2.5.8 (Normalizer of each subgroup of $S_3$, of $Q_8$)

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Exercise 2.5.8 (Normalizer of each subgroup of $S_3$, of $Q_8$)

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