Exercise 1.1.26 (Definition of subgroups)

Question

Assume $H$ is a nonempty subset of $(G, \star)$ which is closed under the binary operation on $G$ and is closed under inverses, i.e., for all $h$ and $k \in H$, $hk$ and $h^{-1} \in H$. Prove that $H$ is a group under the operation $\star$ restricted to $H$ (such a subset $H$ is called a subgroup of $G$).

Richard Ganaye · Accepted Answer

\begin{proof} 
Since $G$ is closed under the operation $\star$, the map
$$
\varphi
\left\{
\begin{array}{ccl}
H 	imes H & 	o & H\
(x,y) & \mapsto & x \star y 
\end{array}
ight.$$
is well defined, and defines an operation $\varphi$ on $H$.

\begin{itemize}
\item For all $x,y,z$ in $G$, $(x \star y) \star z = (x \star y ) \star z)$. A fortiori, since $H \subseteq G$, 
$$\forall (x,y,z) \in H^3,\ (x \star y) \star z = (x \star y ) \star z),$$
so the operation is associative in $H$.

\item By hypothesis, $H 
e \varnothing$. Let $a$ be any element of $H$. Then $a$ has an inverse in $G$, and since $H$ is closed under inverses, $a^{-1} \in H$. Since $H$ is closed under the operation $\star$,
$$1 = a \star a^{-1} \in H.$$
Moreover, $1 \star x = x \star 1 = x$ for all $x \in G$, where $H \subseteq G$, so $1$ is an unity in $H$.

\item Let $x$ be any element of $H$. Then $x$ has an inverse $x^{-1} \in G$, and $x^{-1} \in H$ since $H$ is closed under inverses. 
Moreover $x \star x^{-1} = x^{-1} \star x = 1$, so every element of $H$ has an inverse in $H$.
\end{itemize}
So $H$ is a group under the operation $\star$ restricted to $H$.
\end{proof}

Exercise 1.1.26 (Definition of subgroups)

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Exercise 1.1.26 (Definition of subgroups)

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