The intersection of subgroups is a subgroup

The intersection of any number of subgroups, whether it be countable or uncountable, is itself a subgroup. #m/thm/group For any set ${𝐻 𝑖} 𝑖 \in 𝐼$ of subgroups of a group $𝐺$ , the intersection

𝐾 = ⋂ 𝑖 \in 𝐼 𝐻 𝑖

is itself a subgroup of $𝐺$ .

Proof

Clearly $𝑒 \in 𝐻 𝑖$ for all $𝑖 \in 𝐼$ and therefore $𝑒 \in 𝐾$ . Let $𝑎, 𝑏 \in 𝐾$ . Then $𝑎, 𝑏 \in 𝐻 𝑖$ for all $𝑖 \in 𝐼$ . This implies that $𝑎 𝑏 - 1 \in 𝐻 𝑖$ for all $𝑖 \in 𝐼$ and therefore $𝑎 𝑏 - 1 \in 𝐾$ . Therefore $𝐾$ is a subgroup by One step subgroup test.

Properties

If each subgroup is a normal subgroup, so too is their intersection.

Proof of 1

Let $𝑁 𝑖 ⊴ 𝐺$ for $𝑖 \in 𝐼$ , and let $𝐾 = ⋂ 𝑖 \in 𝐼 𝑁 𝑖$ . From above, $𝐾$ is a subgroup. Now let $𝑔 \in 𝐺$ and $𝑘 \in 𝐾$ . Since each subgroup is normal, $𝑔 𝑘 𝑔 - 1 \in 𝑁 𝑖$ for all $𝑖 \in 𝐼$ , hence $𝑔 𝑘 𝑔 - 1 \in 𝐾$ . Therefore $𝐾$ is a normal subgroup.

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