# Tour: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 $gh \in H.$