The intersection of subgroups is a subgroup
The intersection of any number of subgroups, whether it be countable or uncountable, is itself a subgroup. group
For any set
is itself a subgroup of
Proof
Clearly
for all π β π» π and therefore π β πΌ . Let π β πΎ . Then π , π β πΎ for all π , π β π» π . This implies that π β πΌ for all π π β 1 β π» π and therefore π β πΌ . Therefore π π β 1 β πΎ is a subgroup by One step subgroup test. πΎ
Properties
- If each subgroup is a normal subgroup, so too is their intersection.
Proof of 1
Let
for π π β΄ πΊ , and let π β πΌ . From above, πΎ = β π β πΌ π π is a subgroup. Now let πΎ and π β πΊ . Since each subgroup is normal, π β πΎ for all π π π β 1 β π π , hence π β πΌ . Therefore π π π β 1 β πΎ is a normal subgroup. πΎ