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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.