Exercise 2.4.1 (If $H \leq G$, then $\langle H \rangle = H$)

Question

Prove that if $H$ is a subgroup of $G$ then $\langle H \rangle = H$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $H$ be a subgroup of $G$. By definition,
$$\langle H angle = \bigcap _{K\in {\cal A}} K,\qquad 	ext{where}\quad {\cal A} = \{K \leq G \mid H \subseteq K\}.$$
\begin{itemize}
\item Since $H \subseteq K$ for every $K \in {\cal A}$, we obtain $H \subseteq \bigcap\limits _{K\in {\cal A}} K = \langle H angle,$ so
\begin{align}
H \subseteq \langle H angle.
\end{align}
(The inclusion (1) is always true, even if $H$ is not a subgroup: $\langle H angle$ is the least subgroup of $G$ which contains $H$.)

\item Since $H$ is a subgroup of $G$ which contains itself (i.e. $H\subseteq H$), then $H \in {\cal A}$. Hence $ \bigcap\limits _{K\in {\cal A}} K \subseteq H$, so
\begin{align}
\langle H angle \subseteq H.
\end{align}
By (1) and (2), we obtain $\langle H angle =  H$.

$$H \leq G \Rightarrow  \langle H angle =  H.$$

\end{itemize}
\end{proof}

Exercise 2.4.1 (If $H \leq G$, then $\langle H \rangle = H$)

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Exercise 2.4.1 (If $H \leq G$, then $\langle H \rangle = H$)

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