If d divides the order of a group does the group has a subgroup of order d?

The general answer is no. Consider A4={(1),(12)(34),(13)(24),(14)(23),(123),(124),(132),(134),(142),(143),(234),(243)}. The subgroups of A4 are: A4, <(12)(34)>, <(13)(24)>, <(14)(23)>, <(123)>=<(132)>, <(124)>=<(142)>, <(134)>=<(143)>, <(234)>=<(243)>, {(1),(12)(34),(13)(24),(14)(23)}, {(1)}. The order of A4 is 12, the order of <(12)(34)>, <(13)(24)> and <(14)(23)> is 2, the order of <(123)>=<(132)>, <(124)>=<(142)>, <(134)>=<(143)> and <(234)>=<(243)> is 3, the order of {(1),(12)(34),(13)(24),(14)(23)} is 4, and the order of <(1)> is 1. Clearly there are no subgroups of order 6, but 6 definitely divides the order of A4. The statement is true for all finite abelian groups, and when d is a power of a prime (i.e., when d=pk for a prime p and a non-negative integer k).