Task: Give an example for a group $(B, \circ) $ and a subgroup $U \subsetneq B$ 
Task (no homework): Give an example for a group $(B, \circ) $ and a
  subgroup $U \subsetneq B$.

Can you please check if my solution is fine?
Let the set be $B= \left\{-1,1\right\}$, let's take multiplication, so we have the group $(B, \cdot)$.
All axioms of a group are satisfied, so it's indeed a group.
Let the set U be $U= \left\{1\right\}$, so we already have that $U$ is a subset of $B$, and it's a group too, thus it's a subgroup $(U, \cdot)$

Is it correct like that? 
 A: I'm not sure how formal you want to be. Your example is correct, but you may be asked to prove it.
Here's a more slightly formal approach: Let $B=\{-1,1\}$. Let $\cdot$ stand for the multiplication operator. Then $(B, \cdot)$ is a group as $\cdot$ is associative (by definition of multiplication), the elements in $B$ are closed under multiplication ($1\cdot1 = 1, 1\cdot -1 = -1, -1\cdot 1 = -1, 1 \cdot 1 = 1$), we have an identity element, namely $1$, and we have an inverse for each element: $1\cdot 1 = 1$, $-1 \cdot -1  = 1$.
Suppose we take $H = \{1\}$. We will show $H$ is a subgroup of $B$.
To do so we'll apply a subgroup test. In particular, I'll apply the one step subgroup test which states "Let $G$ be a group and let $H$ be a non-empty subset of $G$. If for all $a,b \in H$, $a \cdot b^{-1} \in H$ then $H$ is a subgroup of $G$."
Well in our case we have a pretty simple subset $H$. Still we need to check what the test states: Do we have a group? Yes, $B$. Do we have a non-empty subset of $B$? Yes, $H$. For all $a,b \in H$ is $a \cdot b^{-1} \in H$? Well, we only have element, namely $1$ and $1$ has it's own inverse (e.g., $1 \cdot 1 = 1$) so let $a = 1$ and $b = 1$, then $b^{-1} = 1$ and $a\cdot b^{-1} = 1 \cdot 1 = 1$ and $1 \in H$. So $H$ is a subgroup of $B$ by the one step subgroup test.
Clearly $H \subset B$ and $H \not = B$. Thus $H$ satisfies $H \subsetneq B$ and $H$ is a subgroup of $B$ by the one step subgroup test.
