If H_1 , H_2 are subgroups of a group G then H_1 . H_2 = { h_1 h_2 \in G, h_1 \in H_1 , h_2 \in H_2 } is a subgroup of G.

False

Discussion: If one of the groups is normal then the above assertion would be true. Suppose h_1 , h_1 ' \in H_1 , h_2 , h_2 ' \in H_2 then consider the elements h_1 h_2 and h_1 ' h_2 ' both of which are members of the set of H_1 H_2 . If H_1 H_2 is a group then their product will also be a member of H_1 H_2 . That is h_1 h_2 h_1 ' h_2 ' \in H_1 H_2 .

Suppose one of the subgroups, say H_1 is normal then h_2 h_1 ' h_2 ^{-1} \in H_1 or there exists h_1 '' \in H_1 such that h_2 h_1 ' = h_1 '' h_2 . Hence h_1 h_2 h_1 ' h_2 ' = h_1 h_1 '' h_2 h_2 ' and this element definitely belongs to H_1 H_2 as h_1 h_1 '' \in H_1 h_2 h_2 ' H_2 . Existence of identity and inverse are easy to prove.