2.2.15

(a) Let $$H$$ be a subgroup of $$G$$ under a law of composition. By the definition of a group, $$H$$ must contain an identity element.

Denote the identity element of $$H$$ as $$E$$ and the identity element of $$G$$ as $$e$$.

Let $$g \in G$$ be any element of $$G$$. By the definition of the identity element we have $$eg=ge=g$$. Since $$H \subseteq G$$ we can take the identity of G and compose it with $$E$$ in order to obtain $$Ee=eE=E$$.

We know that $$H$$ is closed under the law of composition, so by the result above we deduce that $$e \in H$$. Now take the identity of $$H$$ and compose it with $$e$$ to get $$Ee=eE=e$$.

We've shown that $$Ee=eE=E$$ and $$Ee=eE=e$$. Therefore the identity element of $$H$$ exists, and must be the same as the identity element of $$G$$.

a.alternative) Let $$e' \in H$$ denote the identify element in $$H$$. Take $$a \in H$$, by definition $$ae' = e'a = a$$ in $$H$$. Because $$H \subset G$$, $$a, e' \in G$$ as well. From that $$ae' = e'a = a$$ holds in $$G$$ (because the same law of composition is used) and the identity element (in $$G$$) is unique we get $$e' = e$$, where $$e$$ is the identity element in $$H$$.

b) To show that the inverse of $$a$$ in $$G$$ is the same as the inverse in $$H$$ one can use a similar argument as in a.alternative). Take an arbitrary element $$a \in H$$, and let $$a' \in H$$ denote the inverse of $$a$$ in $$H$$, i.e. $$aa' = a'a = e$$. Because $$H \subset G$$, $$a, a' \in G$$ and $$aa' = a'a = e$$ in $$G$$ as well, and because inverses (in $$G$$) are unique $$a' = a^{-1}$$.