MHB Proving a Subset of a Group is a Group

[cbarker1]

Dear Everyone,

I want to show that a subset of a group is still a group by using the subgroup criterion which states that a subset $H$ of a group $G$ is a subgroup if and only if $H \ne \emptyset$ and for all $x,y \in H, xy^{-1}\in H$. I am having trouble how to show that criterion in the following exercise:

"Let $(G,\star)$ be a group and $H$ be a nonempty subset of $G$ such that $H$ is closed under the group operation $\star$ and is closed under inverses. That is, for all $h_{1}$ and $h_{2}$ in $H$, both $h_{1} \star h_{2} \in H$ and $h^{-1} \in H$. Prove that $H$ is a group under the operation $\star$ restricted to $H$."

My attempt:

We know that the $H$ is not the empty set. We know that $h^{-1}\in H$ because $H \subset G$. So let $h_{1}, h_{2} \in H$ be arbitrary. Let $h_{2}=h^{-1}$. Then $h_{1}\star h^{-1} \in H$.

Thanks,
Cbarker1

 

[I like Serena]

Hi Cbarker1,

You have effectively shown the forward direction.
That is, if $H$ is a subgroup, then for all $x,y \in H: xy^{-1}\in H$.
Indeed that follows because $y^{-1}$ is in $H$ as it is closed under inverses.
And $x\star y^{-1}$ is subsequently in $H$ because $H$ is closed under the group operation.

However, to prove the criterion, we also need to prove it in the reverse direction.
So if $H\ne\varnothing$ and for all $x,y \in H: x \star y^{-1}\in H$, does it follow that $H$ is closed under the group operation? And also closed under inverses?

 

[Olinguito]

Hi Cbarker1.

We know that the $H$ is not the empty set. We know that $h^{-1}\in H$ because $H \subset G$. So let $h_{1}, h_{2} \in H$ be arbitrary. Let $h_{2}=h^{-1}$. Then $h_{1}\star h^{-1} \in H$.

Thanks,
Cbarker1

You’re not quite there yet; all you’ve shown is that $h_1\star h_2\in H$ whereas you want to show that $h_1\star h_2^{-1}\in H$. You need one more step: that $h_2\in H$ implies $h_2^{-1}\in H$ (as $H$ is closed under inverses). Now you can apply closure under $\star$ to $h_1$ and $h_2^{-1}$.

To prove the criterion from the reverse assumption, first show that the identity $e$ is in $H$. (Hint: take $h\in H$ (you can do this as $H$ is nonempty) and apply the criterion to $h,h$.) Then, to show that the inverse of any $h\in H$ is in $H$, apply the criterion to $e,h$, and to show that $h_1\star h_2\in H$ for any $h_1,h_2\in H$, apply the criterion to $h_1,h_2^{-1}$.