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 $H_{i}\!$ be an arbitrary collection of subgroups of a group $G\!$ indexed by $i\in I$. Then, $\textstyle \bigcap _{i\in I}H_{i}$ is again a subgroup of $G\!$.

Note that if the collection $I\!$ 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 $H_{i}\!$ be an arbitrary collection of subgroups of a group $G\!$ indexed by $i\in I.$ Let us denote $H=\textstyle \bigcap _{i\in I}H_{i}.$ Here, $e\!$ denotes the identity element of $G.\!$

To prove: We need to show that $H$ is a subgroup. In other words, we need to show the following:

1. $e\in H$
2. If $g\in H$ then $g^{-1}\in H$
3. If $g,h\in H$ then $gh\in H$

Proof: Let's prove these one by one:

1. Since $e\in H_{i}$ for every $i,\!$ $e\in H.$
2. Take $g\in H$. Then $g\in H_{i}$ for every $i\in I.$ Since each $H_{i}\!$ is a subgroup, $g^{-1}\in H_{i}$ for each $i\in I.$ Thus, $g^{-1}\in H.$
3. Take $g,h\in H.$ Then $g,h\in H_{i}$ for every $i,\!$ so $gh\in H_{i}$ for every $i\in I.$ Thus $gh\in H.$

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