S R Wicker in his book states ... $GF(p^m)$ has all sub-fields of order $p^b$ provided $b$ divides $m$.
Qn: In order to have a sub-filed of order $p^b$ shouldn't we have also the constraint: $p^b - 1$ divides $p^m -1$ ? (that is to say, the multiplicative order of a sub-field should divide the multiplicative order of the entire GF)