Proof for the inverse in sub group is as same as inverse in group

Question

I have a question: Let H be the subgroup of G. I want to prove that if for any element that is in the H, the inverse of that element in the group and sub group is the same. I don't know where to start. I know that and know how to prove that "identity of subgroup is same as identity of group", but I don't know how to proof for the inverse element. Thanks.

Answer 1

Let $\;G;$ be a group, $\;N\;$ a subgroup and $\;n\in N\;$ . Since $\;N\;$ is a group in its own right, there exists $\;n^{-1}\in N\;$ . But there is also an inverse for $\;n\;$ in the whole group, say $\;m^{-1}\;$ .

Thus, we have $\;nn^{-1}=1\;$ ...but this happens all the time within $\;G\;$ , since of course$\;N\subset G\;$ . If you look then at this equation in $\;G\;$ , we get

$$nn^{-1}=1=nm^{-1}\implies n^{-1}=m^{-1}\;\;\text{by uniqueness of inverses in}\;\;G$$

Answer 2

Let $h\in H$ and let $h^{-1}$ be the $H$-inverse of $h$. That means that $hh^{-1} = h^{-1}h = e$. But $h^{-1}\in G$, and multiplication of elements on $H$ works the same way regardless of whether we view the elements as elements in $H$ or in $G$ (the point of a subgroup is that it uses the exact same group operation, after all). So as elements in $G$ we also have $hh^{-1} = h^{-1}h = e$, and $h$ has the same inverse in the two grops.