This problem is a little involved. Consider the Sylow 2-subgroups we know that there are 1 or 3 such Sylow subgroups. If there was only one Sylow 2-subgroup then there is nothing to prove because it is invariant under conjugation and so it is normal. Thus, we will assume there are 3 Sylow 2-subgroups. Let be two of these three Sylow subgroups. Notice that and . Since , therefore cannot be less than or equal to . This forces by applying Lagrange's theorem. Since is of index in and it means . Therefore, the normalizer, must contain both and . By Lagrange's theorem it means is a multiple of and a divisor of . The only possibility is and so is not simple.