Why is it important for a group to be closed under an operation? In my undergrad group theory course, we have seen a lot of groups that are closed under a certain operation. I've also noticed a group now myself in my probability theory course with regards to sigma-fields. My question is why is this property of being closed important? What if a group is not closed? What is the implication of that? I know how to go about proving a particular group is closed but I cannot see the intuition behind it. Can anyone clear this up for me please.
 A: By definition a group is a set $G$ and a map $G \times G \rightarrow G$ such that (...). 
So it makes no sense in this context to say "a group is closed" under the operation, this is automatically true. There is no group that "is not closed".
If you have a set $X$ and a set $Y$ such that $X \subseteq Y$ and you can find some operation $p:X \times X  \rightarrow Y$, then $(X,p)$ might be a group or not. If you think this "operation" can be used to define a group $(X,p)$ then of course you must prove that $p(x,y) \in X$ for all $x,y$. If this is false, then the group axioms make no sense, there is no reason to check them at all.
A: The essential feature of a group is that it describes a notion of symmetry: transformations of space that somehow don't change the figure of interest. If $f$ and $g$ both leave something alone, then so does $f \circ g$, so symmetry is automatically closed.
A variation on a group is called an inverse monoid, and its essential feature is a notion of partial symmetry: transformations of space that somehow don't change part of the figure of interest. If $f$ and $g$ both leave the same thing alone, then so does $f\circ g$, but the portion that $f$ leaves alone versus the portion that $g$ leaves alone may differ, and then the portion that $f \circ g$ leaves alone is usually smaller. Inverse monoids have many features that groups have, but simple things like Lagrange's theorem fail to hold (because “cosets” can have different sizes, because we no longer have full symmetry amongst cosets, only partial; you can see how things can go wrong in this answer which has a fairly concrete inverse monoid example).
The relation to your question is an inconvenient way of defining inverse monoids: a collection of bijections with composition only defined when the range of the first matches the domain of the second. This version is called an inductive groupoid (or “category containing only isomorphisms”).
Until one understands full symmetry, that is groups, it is probably too difficult to say anything interesting about partial symmetry, inverse monoids.
A: As it's already been said in the comments: You want the group operation to be a well defined mapping. I will try to illustrate a bit, why:
Consider $\left(\mathbb{C}\setminus\left\{0\right\},\cdot\right)$, the complex plane without zero and multiplication as group operation. Surely this is a group. Another group would be $\left\{x\in\mathbb{C}:\left|x\right|=1\right\}$ with the same multiplication. This is due to the facts that $\left|x\cdot y\right|=\left|x\right|\cdot\left|y\right\|=1$ and $\left|\frac{1}{x}\right|=\frac{1}{\left|x\right|}=1$ for $\left|x\right|=\left|y\right|=1$.
In practice you often choose the set of a group according to certain properties wich you might use to prove other properties, for instance, if we consider $\left(\left\{x\in\mathbb{C}:\left|x\right|=1\right\},\cdot\right)$ again, we can see that the group operates bounded, even non-expansively on $\mathbb{C}$, which are very useful properties in some cases.
If you would for instance take $\left(\left\{x\in\mathbb{C}:\frac{1}{2}\leq\left|x\right|\leq2\right\},\cdot\right)$ you would still have that the inverse of each element is a member of the set, you can also easily see that the "group" operates bounded on $\mathbb{C}$, that means $\forall y\in\mathbb{C}, \forall \frac{1}{2}\leq\left|x\right|\leq2$ you have that $\left|xy\right|\leq2\left|y\right|$ but if you iteratively apply a sequence of "group" operations $x_{1},...,x_{n}$ on $y\in\mathbb{C}$ you cannot say that $\left|x_{1}...x_{n}y\right|\leq2\left|y\right|$. This is because you do not have that $x_{1}\cdot...\cdot x_{n}$ is a member of your "group".
I hope that helps :)
