Connection between principal ideal and Cayley's theorem (?) A (left) principal ideal $I$ of $(R,+,\times)$ is a subset of $R$ of the form $Ra = \{ra : r \in R\}.$ So, on the multiplication table, the principal ideal is a column (or a row) in the multiplication of the element $a$ with all the other elements of the ring, $R.$ 
My understanding is the ideal has to form a subgroup under addition.
On the other hand, and within a finite (?) group $G$, I think it's Cayley's theorem that guarantees that each entry in the multiplication table (each column or row) includes a permutation of all the elements in the group.

I'm asking for help understanding why, then, not all principal ideals contain exactly all the elements in the ring.

Evidently, this monotonous and predictable result wouldn't be of much interest, and I presume the answer is likely to lie in the fact that the set $R$ forms a group under addition, but not under multiplication. Is this the reason?
 A: I think you may have just a simple misunderstanding of some basic definitions.  Consider a simple example to see if that clears things up:
$$\langle 2 \rangle = \{0, \pm 2, \pm 4, \ldots \}$$
in the ring $\langle \mathbb{Z}, +, \cdot, 0,1\rangle $
Clearly the thing above is a principal ideal but it also is not the entire set of all integers.  Hopefully that helps.
A: Note that a ring $(R, +, \cdot)$ forms an abelian group over addition (that is, $(R, +)$ is an abelian group. Now depending on your definition, $(R, \cdot)$ is considered as either a semigroup or a monoid. (Informally, a semigroup is a group that doesn't have identity and isn't closed under inverses. A monoid is a semigroup with identity). The important point is that not every element of $R \setminus \{0\}$ necessarily has a multiplicative inverse.
The group structure, namely closure under inverses, provides for the cancellation law. In this manner, the natural left action induces a bijection. This is basically the result of Cayley's Theorem.
As not every element of $R \setminus \{0\}$ has a multiplicative inverse, the ring multiplication operation need not induce a bijection. Consider the ring $\mathbb{Z}_{6}$. We have that $2 \cdot 2 \equiv 2 \cdot 5 \equiv 4 \pmod{6}$. Clearly, $2$ and $5$ are distinct elements in $\mathbb{Z}_{6}$. So while $(2)$ is clearly a principal ideal in $\mathbb{Z}_{6}$, $(2)$ is a proper ideal of $\mathbb{Z}_{6}$. That is, $(2) \neq \mathbb{Z}_{6}$.
