# Thread: Prove H is a subgroup

1. ## Prove H is a subgroup

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

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

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

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

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

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

Is this correct?

2. ## Re: Prove H is a subgroup

can you write the original problem ? its little confusing since its not clear which are the 'givens'

3. ## Re: Prove H is a subgroup

Originally Posted by issacnewton
can you write the original problem ? its little confusing since its not clear which are the 'givens'
Everything to the left of the word "prove" is given.

4. ## Re: Prove H is a subgroup

Originally Posted by dwsmith
[tex]
Since $h,k\in H$ and $k^{-1}\in H$, $h*(k^{-1})^{-1}=h*k=hk\in H$
since $hk \in H$ (given), then are you trying to show again that $hk \in H$ ? $\smile$

5. ## Re: Prove H is a subgroup

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

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

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

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

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

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

Is this correct?
What you did is either irrelevant or wrong, except for the associative part.

You already know that H is closed under * and that any h in H has inverse in H. So all you have to do is to prove that e is in H. just multiply h with h^-1.