Intersection of subgroups is subgroup

This article gives the statement, and possibly proof, of a basic fact in group theory.
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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, ${\bigcap }_{i\in I}{H}_i$ is again a subgroup of $G$.

Proof

Given: Let $H_i$ be an arbitrary collection of subgroups of a group $G$ indexed by $i\in I$. Let us denote $H={\bigcap }_{i\in I}{H}_i$.

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:

• Since $e\in H_i$ for every $i$, $e\in H$

1. 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, $g^{-1}\in H$.
2. 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 $gh\in H$.