Prime order not implies simple for algebra loops

Statement
It is possible to have an algebra loop that is a loop of prime order (i.e., its order is a fact about::prime number) but has a proper nontrivial fact about::normal subloop, and is hence not a simple loop.

Proof
Consider the algebra loop of order five with multiplication table:

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

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

It is easy to check that the subset $$\{ 1,2 \}$$ is a normal subloop of the loop. In fact, it is precisely the center of the loop, i.e., its elements commute and associate with everything.