Can a group with elements $I,C,L,X$ have $CL$ as an entry in its Cayley table? I am attempting to understand a certain phenomenon (which I do not currently understand well enough that explaining it would add value to the question) for which I have reasoned out the following would-be Cayley table:  \begin{array}{c|cccc}\textbf{∘} & \textbf{I} & \textbf{C} & \textbf{L} & \textbf{X}
\\ \hline \textbf{I} & \text{I} & \text{C} & \text{L} & \text{X}
\\ \textbf{C} & \text{C} & \text{I} & \text{} & \text{I}
\\ \textbf{L} & \text{L} & \text{} & \text{I} & \text{}
\\ \textbf{X} & \text{X} & \text{I} & \text{} & \text{I}\end{array}
The issue is that the four blank spaces, while all equal to each other, are not equal to $C$, $L$, $X$, or the identity $I$, so $\{I,\ C,\ L,\ X\}$ is not closed under $∘$.  My idea is to introduce into the set an extra element arbitrarily named $CL$, defined as $C ∘ L$ (or $L ∘ C$ or $L ∘ X$ or $X ∘ L$).  This leads to the following Cayley table:  \begin{array}{c|cccc}\textbf{∘} & \textbf{I} & \textbf{C} & \textbf{L} & \textbf{X} & \textbf{CL}
\\ \hline \textbf{I} & \text{I} & \text{C} & \text{L} & \text{X} & \text{CL}
\\ \textbf{C} & \text{C} & \text{I} & \text{CL} & \text{I} & \text{L}
\\ \textbf{L} & \text{L} & \text{CL} & \text{I} & \text{CL} & \text{C}
\\ \textbf{X} & \text{X} & \text{I} & \text{CL} & \text{I} & \text{L}
\\ \textbf{CL} & \text{CL} & \text{L} & \text{C} & \text{L} & \text{I}\end{array}
which appears to be an abelian group.
$1)$  Is this a valid way to build a group, even though $CL$ is not an "independent" element of the original set, but a composition of two elements artificially treated as a new element?
$2)$  I haven't studied Abstract Algebra in any depth, so I'm not quite sure what the obvious thing to do with a group is once you've built it.  Without knowing what phenomenon I actually built this group to model, is it possible to make inferences about what it represents from its Cayley table?  What known applications, if any, are modeled by this group?
 A: In general, sure, you can say "the product of $C$ and $L$ is not one of the four elements I already have, so I must expand my multiplication table to include it". This is perfectly reasonable, as long as you make sure that the expansion respects the group axioms.
In your case, however, what you have is not a group. The first and most obvious issue to me is that there are rows and columns where $I$ appears twice. This cannot happen in a group: each row and each column of the multiplication table must list each element in the group exactly once each.
And if you didn't have an application in mind when constructing the group, it's somewhat difficult to say what it could be used for, particularly since it isn't a group. A particular group is often only as useful as its applications. That being said, many groups appear in several places, as there are many different places where the same permutations and symmetries may appear. This is easier to discuss with concrete examples, though.
A: As the other answers say, you have not in fact produced a group. However, you also ask whether the general idea of adding elements to "fill gaps" is valid; let me say a bit about how to formalize that idea. (If you're already familiar with group presentations, skip to the last section of this answer.)
After reading this answer, you should calculate exactly what group this construction - together with the "abelianization rules," if you want an abelian group at the end, which I think you do - yields when fed your starting structure.

The first key notion is that of a free group. Given a set $X$, the free group on $X$ (denoted "$F_X$") is basically the "default" way to fit $X$ into a group: elements of the group are simply formal sequences of the form $$x_1^{n_1}x_2^{n_2}...x_k^{n_k},$$ where

*

*each $x_i$ is in $X$,


*each $n_i$ is in $\mathbb{Z}\setminus 0$, and


*we never have $x_i=x_{i+1}$,
and the group operation is given by concatenation together with "combining like terms," so e.g. for $a,b,c,d\in X$ we have $$(a^2d^{-3})*(d^4b^{-5}c^1)=a^2d^1b^{-5}c^1$$ and $$(a^2d^{-1})(d^1b^1)=a^2b^1$$ (note that taking $k=0$ corresponds to the empty string - this gives our group an identity element).

The second notion, which builds on the first, is that of a group presentation. Intuitively, a group presentation consists of a set of generators together with a set of rules which tell us when various expressions in terms of those generators are equal to each other. For example, we can get the abelian version of the free group in this way: the free abelian group on $X$ is just $F_X$ modified by the rules $$x_1^{n_1}x_2^{n_2}=x_2^{n_2}x_1^{n_1}$$ for each $x_1,x_2\in X$. We call these rules relations or relators.
The precise definition is this. Suppose we have a set $\mathcal{P}$ of pairs of elements of $F_X$. The set $\mathcal{P}$  determines a normal subgroup $NS(\mathcal{P})$ of $F_X$, namely the intersection of all normal subgroups of $F_X$ containing each element of the form $$t_1t_2^{-1}$$ for $\langle t_1,t_2\rangle\in\mathcal{P}$. The group we're building is then just $F_X/NS(\mathcal{P})$.

*

*Note that this means that presentations are really just new language for an old concept. However, they're often more intuitive to work with, they are interesting objects in their own right, and they make sense in contexts more general than groups where "quotient objects" are more complicated. Re: this last point, basically a presentation is a description of a congruence relation, and when we don't have inverses we can't in general reduce a congruence relation (which is a set of pairs of objects) to a set of individual objects (in the group context, the kernel of the corresponding quotient map).


So here's how we put these together in this context. Say I have a set $X$ together with a partial binary operation $*$ on $X$. We can consider the free group on $X$ subject to the relations $$x_1x_2=x_3$$ for each $x_1,x_2,x_3\in X$ such that $x_1*x_2$ is defined and equal to $X$ (and if we want abelian-ness, we can throw in the commutativity rules above as well). This will always yield a group and is perfectly well-defined.
However, we don't always get what we'd expect. Consider for example the case where we have $X=\{a,b\}$ and let $*$ be the operation on $X$ given by $$a*a=a*b=b*a=a,\quad b*b=\mbox{undefined}.$$ When we run the construction above, things collapse: in any group where $a*b=a*a$ we must have $a=b$, and so in fact the group $$F_X/\langle \{aaa^{-1}, aba^{-1}, baa^{-1}\}\rangle$$ (which is intuitively "the group build from $(X,*)$") is just the trivial group. Basically, the construction above doesn't just "fix" undefinedness issues, it also "fixes" all the other "anti-groupinesses" of $*$, and this can sometimes have huge impacts.
George Bergman wrote a lovely paper (here, with corrections and updates here) which goes into detail on this topic in the context of rings as opposed to groups. I highly recommend it once you're familiar with the notion of quotient rings.
A: What you have created is not a group, since it violates the Latin square property.
