Fully invariant not implies normal in loops

Statement
It is possible to have a fully invariant subloop of a loop that is not a normal subloop.

Related facts

 * Fully invariant not implies Lagrange-like
 * Characteristic not implies normal in loops
 * Normal implies Lagrange-like

Proof
Let $$L$$ be a loop with elements $$1,2,3,4,5$$ and operation as follows:

In other words, this is the algebra loop corresponding to the Latin square:

$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 1 & 5 & 3 & 4 \\ 3 & 5 & 4 & 2 & 1 \\ 4 & 3 & 1 & 5 & 2 \\ 5 & 4 & 2 & 1 & 3 \\\end{pmatrix}$$

Suppose $$S$$ is the subloop $$\{ 1,2 \}$$ of $$L$$. Then:


 * $$S$$ is a fully invariant subloop of $$L$$: In fact, it is the only proper nontrivial subloop, and also the only subloop of order two, so its image under any endomorphism must be either itself or the trivial subloop.
 * $$S$$ is not a normal subloop of $$L$$: For instance, $$(S * 3) * 3 \ne S * (3 * 3)$$.