Fully invariant not implies normal in loops

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This article gives the statement and possibly, proof, of a non-implication relation between two subloop properties. That is, it states that every subloop satisfying the first subloop property (i.e., fully invariant subloop) need not satisfy the second subloop property (i.e., normal subloop)
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Statement

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

Related facts

Proof

Further information: non-power-associative loop of order five

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

* 1 2 3 4 5
1 1 2 3 4 5
2 2 1 5 3 4
3 3 5 4 2 1
4 4 3 1 5 2
5 5 4 2 1 3

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).