Let $$\langle H_1, *\rangle$$​ and $$\langle H_2, *\rangle$$​ be subgroups of group $$\langle G, *\rangle.$$​ Which of the following is False?

Correct option is A

We are given two subgroups $$H_1$$​ and $$H_2$$​ of a group G. We need to identify the false statement. Let:
​$$H_1H_2=\{h_1*h_2:h_1\in H_1,\ h_2\in H_2\}$$​​
The important point is that the product of two subgroups need not itself be a subgroup unless an additional condition is satisfied, such as $$H_1H_2=H_2H_1.$$​​
(a) $$H_1H_2$$​ is a subgroup — False
In general,
​$$H_1H_2=\{h_1h_2:h_1\in H_1,h_2\in H_2\}$$​​
is not necessarily a subgroup of G.
For $$H_1H_2$$​ to be a subgroup, a sufficient condition is:
​$$\boxed{H_1H_2=H_2H_1}$$​​
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) $$H_1H_2$$​ is a normal subgroup if $$H_1,H_2$$​ are normal — True
If:
​$$H_1\trianglelefteq G \quad\text{and}\quad H_2\trianglelefteq G,$$​​
then their product
​$$H_1H_2$$​​
is a subgroup of G.
Moreover, it is normal in G.
For $$g\in G:$$​​
​$$g(H_1H_2)g^{-1} = (gH_1g^{-1})(gH_2g^{-1})$$​​
Since $$H_1$$​ and $$H_2$$​ are normal:
​$$gH_1g^{-1}=H_1$$​​
and
​$$gH_2g^{-1}=H_2$$​​
Therefore:
​$$g(H_1H_2)g^{-1}=H_1H_2.$$​​
Hence:
​$$\boxed{H_1H_2\trianglelefteq G}$$​​
So (b) is true.
(c) $$H_1\cap H_2$$​ is a subgroup — True
The intersection of any two subgroups of a group is always a subgroup.
Since:
​$$H_1\leq G,\qquad H_2\leq G,$$​​
we have:
​$$\boxed{H_1\cap H_2\leq G}$$​​
The identity element belongs to both subgroups, and closure and inverses are preserved in the intersection.
Thus, (c) is true.

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