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**dwsmith** $\displaystyle H\neq\emptyset$, $\displaystyle H\subseteq (G,*)$, and G is close under inverses $\displaystyle h,k\in H$ and $\displaystyle hk,h^{-1}\in H$. Prove $\displaystyle H\leq G$

Since G is a group under $\displaystyle *$, H is associative under $\displaystyle *$

$\displaystyle h^{-1}*(hk)=(h^{-1}h)*k=e*k=k\in H$ and $\displaystyle e\in H$

$\displaystyle e=hh^{-1}=h^{-1}h\Rightarrow h^{-1}\in H$

Since $\displaystyle h,k\in H$ and $\displaystyle k^{-1}\in H$, $\displaystyle h*(k^{-1})^{-1}=h*k=hk\in H$

Therefore, since H is closed, associative, has the identity, and has inverses, $\displaystyle H\leq G$

Is this correct?