How can we explain the logic behind the closure axiom in Group Theory to a non-math major? What is the reasoning for why it is necessary? If the result of compounding two elements falls outside the group, why does that matter?
Why is Closure a necessary axiom in Group Theory?
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There is no logic to it. It's just how we define groups:

A magma just obeys closure, and is the simplest possible algebraic structure. Otherwise, you would basically just have a set (since the operation is meaningless). If I take the set $\{1,2,3\}$ and apply addition, there is nothing inherently special about addition to this set. I could just as easily apply multiplication. Closure is just a simple relation to link the set to the operation that closes it.
As for why groups in particular obey closure, it's simply because they are defined as a magma with particular properties (and any other combination just has a similar label). i.e groups are nothing special in terms of requiring closure, it's just a name we associate for structures that have the property of closure among other requirements
On an additional note, it's worth pointing out that the entire concept of algebraic structures is to generalise existing concepts. i.e since there are groups of cycles or matrices, if I prove something in general about a group, it will be true for both cycles and matrices. Closure is a common thing that most useful algebra (such as matrices and cycles) obey. Hence it would make sense for it to be a property of groups (so we can prove more lemmas, theorems etc).
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If the result of compounding two elements falls outside the group, why does that matter?
Where does the compound of two elements lie then?
You may be assuming the group lies inside some ambient space, so that the compound of two elements would still lie "somewhere", just not inside the group. This may happen in concrete examples of groups you have seen. However, when you stick strictly to the axioms of a group, no data around "ambient space" is provided. So the best you can do is to then just declare that sometimes, the compound of two elements is undefined.
This declaration would bring us a lot of other headaches. Groups think about compounding one after another. If sometimes $b * c$ does not make sense, now whenever you think about $a * (b * c)$, you need to make sure you don't fall into the undefined scenario. This is sometimes possible (for example, avoiding division by zero in real numbers), but is hard to describe precisely in abstract. So now you have two ways out:
- Either you work with a more concrete situation, where you can describe more precisely when compounding is not possible,
- or you just restrict your attention to elements where compounding is always possible; but this exactly means you are accepting the axiom of closure!
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Often times, a group is meant to model some form of "symmetry." For example, if you think about a rubik's cube, basically each "move" of the rubiks's cube can be represented as an element of the group, and this group is the set of operations that act on the rubik's cube. The group has to be closed because each move will result in another possible configuration of the rubik's cube; if the group were not closed, then this means that there are "invalid moves," i.e. one that results in an impossible configuration of the cube.
For another example, a group might be the set of symmetries of a square (all rotations and reflections). Because the symmetries do not "change" the square (after applying a 90 degree rotation, for example, the square looks the same as it started), we require that the group be closed.
If an operation that is defined over a set is not closed, you can't really calculate within this set. Here is a example:
Let the set $S$ contain these two elements:
$$ S=\{, \} $$
and let's define an operation $◇$ that is not closed, so that the link table is incomplete:
So we have:
Then $(◇)◇(◇)$ would also be an undefined expression, but in a group this expression is expected to be an element of $S$.
Now let's define it with something else, for example this:
Then we no longer have undefined places in the link table:
But what would then be the result of (◇)◇? The expression inside the brackets gives , but is not a member of $S$, and therefore ◇ is not defined.
You can't do all the things you usually could do with groups if the operation is not closed.