Proving a subgroup of a group My question states, Let H be subgroup of a group G. Prove that HH=H and that $H^{-1}=H$  I started the 1st proof by saying  HH=H  We will show that $HH\subseteq H$  Then $x\in HH$  Then $x=ab$ where $a\in H$ and $b\in H$  I am lost right now trying to prove HH=H and $H^{-1}=H$  Anybody can help me?
I already mentioned in my comment that $HH=H$. 

We know that $H&lt;G$. We have $$H^{-1}=\{h^{-1}: h\in H\}.$$

Let $x\in H^{-1}$. Then $x=h^{-1}$ for some $h\in H$. Because $H&lt;G$, we have $h^{-1}\in H$. Thus, $x\in H$. Hence, $H^{-1}\subset H$.

Let $y\in H$. Then $y^{-1}\in H$. Hence, $y=(y^{-1})^{-1}\in H^{-1}$. Thus, $H\subset H^{-1}$.

Done.Hope it helps.
Given   that   H   be   a   sub   group   of   G
So   we   have to   show   that   HH   =   H
For that it is enough to show that HH ⊆ H and H ⊆ HH
Now   let   h_1  h_2∈HH        where   h_1∈ H,   h_2∈ H
∴   h_1∈H ,h_2∈ H   ⇒  h_1  h_2∈H      (  ∵ closed law  )
∴   h_1  h_2 ∈ HH   ⇒   h_1  h_2 ∈ H
⇒   HH⊆ H    ------------ (1)
Again    let   h ∈ H   then   we   have to   show   that   h ∈ HH
Let   h ∈ H   and   e ∈ H   ⇒ he ∈ HH
                                             ⇒ h ∈ HH

∴      h ∈ H  ⇒   h ∈ HH 
                         ⇒  H ⊆ HH   ------------------------- (2)
∴      from   (1)  and  (2)  we  get  HH =  H ( ∵  HH ⊆ H and  H ⊆ HH)
Hence  proved