If you find any mistakes, please make a comment! Thank you.

Subsets of a group 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 $H$.


Solution: 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.

(1) $\overline{\star}$ is a binary operator on $H$ by hypothesis.

(2) $\overline{\star}$ is associative because $\star$ is associative.

(3) 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 $1_H = 1_G$.

(4) 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 $\star$.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu