Question

(b) Let {H_i} be any collection of subgroups of G. Prove that $\cap H_i$is a subgroup of G.

Short answer

We proved that if {H_i} is a collection of subgroups of G, then $\cap H_i$is a subgroup of G.

Step-by-step solution

Subgroups of a group G

We have the definition of subgroup that,

Let (G,*) be a group under binary operation, say *,A non-empty subset H of is said to be a subgroup of G, if (H,*) is itself a group.

Suppose {H_i} is a collection of subgroups of G.

To show that ∩Hi is a subgroup of G

Let {H_i} be a collection of subgroups of G, $\forall \mathrm{i}\in \mathrm{I}$, where I is a family of index.

Now,

$$\begin{gathered} \underset{\mathrm{i}\in \mathrm{I}}{\cap }{\mathrm{H}}_{\mathrm{i}}\ne \mathrm{\varphi },\mathrm{since}1\in {\mathrm{H}}_{\mathrm{i}},\forall \mathrm{i}\in \mathrm{I} \\ \mathrm{Let}\mathrm{a},\mathrm{b}\in \underset{\mathrm{i}\in \mathrm{I}}{\cap }{\mathrm{H}}_{\mathrm{i}} \\ \mathrm{a},\mathrm{b}\in {\mathrm{H}}_{\mathrm{i}},\forall \mathrm{i}\in \mathrm{I} \\ {\mathrm{ab}}^{-1}\in {\mathrm{H}}_{\mathrm{i}},\forall \mathrm{i}\in \mathrm{I}...(\mathrm{Since},\{{\mathrm{H}}_{\mathrm{i}}\}\mathrm{is}\mathrm{a}\mathrm{subgroup}) \\ {\mathrm{ab}}^{-1}\in \underset{\mathrm{i}\in \mathrm{I}}{\cap }{\mathrm{H}}_{\mathrm{i}} \end{gathered}$$

Therefore, By subgroup criterion,

$\underset{\mathrm{i}\in \mathrm{I}}{\cap }{\mathrm{H}}_{\mathrm{i}}$is a subgroup of G.