- #1

pauloromero1983

- 2

- 0

My question is how to arrive to such conclusion. I am aware that, for every ordered pair of \(G\times G\) there's \(n\) images (since \(X\) was assumed to have \(n\) elements). For a concrete example, let be \(G\) a group of 2 elements. Then, there are 4 ordered pairs. Each pair has 2 images, so the total number of maps would be 4*2=8. However, by use of the relation \(n^{n^{2}}\) we get \(2^{2^{2}}=16\), i.e, there are 16 different maps, not 8. I am missing something here, but I don't know what exactly what the error is.