Given the group $(G,\circ)$ why is a subset of $G$ together with $\circ$ associative as well? I'm trying to proof that the subset $H$ of $G$ together with $\circ$ is also a group i.e. a subgroup and I'm stuck in the proof that $(H,\circ)$ is associative. 
 A: Let $a$, $b$, and $c$ be in $H$.  Then since $H$ is a subset of $G$, we know $a$, $b$, and $c$  are in $G$ and satisfy the associative law:  $a\circ(b\circ c) = (a\circ b)\circ c$.  Therefore $\circ$ is associative for $H$.
Now, not every subset $H$ of $G$ is a subgroup, but your question seems to imply you've got the other criteria figured out.
A: By definition, a subset $H$ of a group $G$ is a subgroup when $H$ is a group using the operation of $G$.
In principle, this would require you to check all group properties for $H$:
$H$ is closed under the operation, there is a neutral element, there exists an inverse for each element in $H$, and that the operation is associative.
But since the operation of $H$ is the same as the operation of $G$, just restricted to $H$, it inherits its properties and you only have to check that:


*

*$H$ is closed under the operation

*$H$ contains the neutral element of $G$ — you don't need to prove that it is a neutral element for $H$

*$H$ contains the inverse in $G$ of each of its element — you don't need to prove that this inverse is an inverse in $H$

*You don't need to prove that the operation is associative in $H$ because it is associative in $G$ 
