Tour:Intersection of subgroups is subgroup
This article adapts material from the main article: intersection of subgroups is subgroup
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PREREQUISITES:Definition of group and subgroup (preferably, the universal algebraic definitions)
WHAT YOU NEED TO DO:
- Try proving that in any group, an intersection of subgroups is again a subgroup. Here, by intersection we mean the intersection of the underlying subsets.
- Check out the proof below after you've tried.
Statement
Verbal statement
The intersection of any arbitrary collection of subgroups of a group is again a subgroup.
Symbolic statement
Let be an arbitrary collection of subgroups of a group
indexed by
. Then,
is again a subgroup of
.
Note that if the collection is empty, the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.
Proof
Given: Let be an arbitrary collection of subgroups of a group
indexed by
Let us denote
Here,
denotes the identity element of
To prove: We need to show that is a subgroup. In other words, we need to show the following:
-
- If
then
- If
then
Proof: Let's prove these one by one:
- Since
for every
- Take
. Then
for every
Since each
is a subgroup,
for each
Thus,
- Take
Then
for every
so
for every
Thus
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