**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.

*Related*

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