Exercise 2.4.4 (If $H \leq G$, then $\langle H - \{1\} \rangle = H$)

Question

Prove that if $H$ is a subgroup of $G$ then $H$ is generated by the set $H - \{1\}$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $H$ be a subgroup of $G$.
\begin{itemize}
\item First $H - \{1\} \subseteq H$.
\item If $K$ is any subgroup of $G$ which contains $H - \{1\}$, since $1 \in K$, then $H = (H - \{1\}) \cup \{1\} \subseteq K$, so $H \subseteq K$
\end{itemize}
This shows that $H$ is the smallest group (for inclusion) which contains $H - \{1\}$, so
$$\langle H \setminus \{1\} \rangle = H.$$
(This is true even if $H = \{1\}$, because $\langle \varnothing \rangle = \{1\}$.)
\end{proof}

Exercise 2.4.4 (If $H \leq G$, then $\langle H - \{1\} \rangle = H$)

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Exercise 2.4.4 (If $H \leq G$, then $\langle H - \{1\} \rangle = H$)

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