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Let ⟨H1,∗⟩\langle H_1, *\rangle⟨H1​,∗⟩​ and ⟨H2,∗⟩\langle H_2, *\rangle⟨H2​,∗⟩​ be subgroups of group ⟨G,∗⟩.\langle G, *\rangle.⟨G,∗⟩.​
Question

Let ⟨H1,∗⟩\langle H_1, *\rangle​ and ⟨H2,∗⟩\langle H_2, *\rangle​ be subgroups of group ⟨G,∗⟩.\langle G, *\rangle.​ Which of the following is False?

A.

H1.H2={h1∗h2:h1∈H1 and h2∈H2}H_1.H_2=\{h_1*h_2:h_1\in H_1\text{ and }h_2\in H_2\}​ is a subgroup.

B.

H1.H2={h1∗h2:h1∈H1 and h2∈H2}H_1.H_2=\{h_1*h_2:h_1\in H_1\text{ and }h_2\in H_2\}​ is a normal subgroup given that H1H_1​ and H2H_2​ are normal subgroups.

C.

⟨H1∩H2,∗⟩\langle H_1\cap H_2,*\rangle​ is a subgroup.

D.

⟨H1∪H2,∗⟩\langle H_1\cup H_2,*\rangle​ is a subgroup.

Correct option is A

We are given two subgroups H1H_1​ and H2H_2​ of a group G. We need to identify the false statement. Let:
​H1H2={h1∗h2:h1∈H1, h2∈H2}H_1H_2=\{h_1*h_2:h_1\in H_1,\ h_2\in H_2\}​​
The important point is that the product of two subgroups need not itself be a subgroup unless an additional condition is satisfied, such as H1H2=H2H1.H_1H_2=H_2H_1.​​
(a) H1H2H_1H_2​ is a subgroup — False
In general,
​H1H2={h1h2:h1∈H1,h2∈H2}H_1H_2=\{h_1h_2:h_1\in H_1,h_2\in H_2\}​​
is not necessarily a subgroup of G.
For H1H2H_1H_2​ to be a subgroup, a sufficient condition is:
​H1H2=H2H1\boxed{H_1H_2=H_2H_1}​​
This condition is automatically satisfied if, for example, one of the two subgroups is normal.
Since the question does not provide such a condition in statement (a), the statement is false.
(b) H1H2H_1H_2​ is a normal subgroup if H1,H2H_1,H_2​ are normal — True
If:
​H1⊴GandH2⊴G,H_1\trianglelefteq G \quad\text{and}\quad H_2\trianglelefteq G,​​
then their product
​H1H2H_1H_2​​
is a subgroup of G.
Moreover, it is normal in G.
For g∈G:g\in G:​​
​g(H1H2)g−1=(gH1g−1)(gH2g−1)g(H_1H_2)g^{-1} = (gH_1g^{-1})(gH_2g^{-1})​​
Since H1H_1​ and H2H_2​ are normal:
​gH1g−1=H1gH_1g^{-1}=H_1​​
and
​gH2g−1=H2gH_2g^{-1}=H_2​​
Therefore:
​g(H1H2)g−1=H1H2.g(H_1H_2)g^{-1}=H_1H_2.​​
Hence:
​H1H2⊴G\boxed{H_1H_2\trianglelefteq G}​​
So (b) is true.
(c) H1∩H2H_1\cap H_2​ is a subgroup — True
The intersection of any two subgroups of a group is always a subgroup.
Since:
​H1≤G,H2≤G,H_1\leq G,\qquad H_2\leq G,​​
we have:
​H1∩H2≤G\boxed{H_1\cap H_2\leq G}​​
The identity element belongs to both subgroups, and closure and inverses are preserved in the intersection.
Thus, (c) is true.

​

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