Algebra Solutions: In a group, subsets which are closed under multiplication and inversion are groups

Let $(G, \star)$ be a group. Show that a nonempty subset $H \subseteq G$ which is closed under $\star$ and inversion is a group under the restriction $\overline{\star}$ of $\star$ to $H$ .

We need to demonstrate that $\overline{\star}$ is a binary operator on $H$ , that $\overline{\star}$ is associative, that an identity element $e \in H$ exists, and that every element has an inverse.

$\overline{\star}$ is a binary operator on $H$ by hypothesis.
$\overline{\star}$ is associative because $\star$ is associative.
Since $H$ is nonempty, there exists some $h \in H$ . Since $H$ is closed under inversion in $G$ , $h^{-1} \in H$ . And since $H$ is closed under $\star$ , we have $h \star h^{-1} = 1_G \in H$ . Moreover, for all $x \in H$ we have $x \overline{\star} 1_G = x \star 1_G = x$ ; thus $H$ has an identity element and in fact $1_H = 1_G$ .
Let $x \in H$ . Then since $H$ is closed under inversion in $G$ we have $x^{-1} \in H$ , and $x \overline{\star} x^{-1} = x \star x^{-1} = 1_G = 1_H$ . Thus every element of $H$ has an inverse.

So $H$ is indeed a group under the restriction of $\star$ .

Algebra Solutions

In a group, subsets which are closed under multiplication and inversion are groups

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