Multiplication/Division in Galois Field (2^8)
In GF(28), 7 &times; 11 = 49.  The discrete logarithm trick works just fine.

Your mistake is in assuming that Galois field multiplication works the same way as normal integer multiplication.  In prime-order fields this actually is more or less the case, except that you need to reduce the result modulo the order of the field, but in fields of non-prime order the multiplication rules are different.

Let's do 7 &times; 11, for example.  Noting that 7 = 1112 and 11 = 10112, we can calculate 7 &times; 11 in binary as:

    111 ×
   1011 =
   ------
    111 +
   1110 +
 111000 =
 --------


So far, everything works the same as in ordinary integer multiplication.  But whereas in the integers we'd propagate carries while doing the addition, and thus end up with

&nbsp; &nbsp; &nbsp; 1112 + 11102 + 1110002 = 101012 + 1110002 = 10011012 = 77,

in GF(2n) addition is done bitwise without carries (i.e. GF(2n) addition is the same as bitwise XOR), and thus we get

&nbsp; &nbsp; &nbsp; 1112 + 11102 + 1110002 = 10012 + 1110002 = 1100012 = 49.

(Of course, if the result exceeded the group order, we'd also have to reduce it modulo the reduction polynomial, but in this case that doesn't happen for n &ge; 7.)