Let G be a group and X 
⊂
 ﻿G a subset 
(
not necessarily a subgroup
)
 ﻿of G
.
 ﻿Let H be the
intersection of all subgroups of G that contain X
.
 ﻿Show that H is a subgroup of G
.
 ﻿It is
called the subgroup of G generated by the subset X
.

To show that \( H \) is a subgroup of \( G \), we will use the definition of \( H \) and apply the subgroup criteria. The set \( H \) is defined as the **intersection of all subgroups of \( G \) that contain \( X \)**. We need to demonstrate that \( H \) satisfies the subgroup properties.

### Step-by-Step Solution

#### 1. **Definition of \( H \)**:
Let \( H = \bigcap_{\alpha} H_\alpha \), where each \( H_\alpha \) is a subgroup of \( G \) that contains \( X \). In other words, \( H \) consists of all elements of \( G \) that are in every subgroup \( H_\alpha \) containing \( X \).

#### 2. **Subgroup Criterion**:
To show that \( H \) is a subgroup of \( G \), we need to verify that it satisfies the following three conditions:
1. \( H \) is **non-empty**.
2. \( H \) is **closed under the group operation**.
3. \( H \) is **closed under taking inverses**.

#### 3. **Non-emptiness**:
Since \( G \) is a group, it contains the identity element \( e \). Every subgroup of \( G \) must contain \( e \), because the identity is a required element of any group. Hence, each \( H_\alpha \) contains \( e \), so \( e \in H \). Therefore, \( H \) is non-empty because it contains at least the identity element \( e \).

#### 4. **Closed under the group operation**:
Let \( a, b \in H \). By the definition of \( H \), this means that \( a, b \in H_\alpha \) for every subgroup \( H_\alpha \) containing \( X \). Since each \( H_\alpha \) is a subgroup, and subgroups are closed under the group operation, \( ab \in H_\alpha \) for every \( H_\alpha \). Therefore, \( ab \in H \). Thus, \( H \) is closed under the group operation.

#### 5. **Closed under taking inverses**:
Let \( a \in H \). By the definition of \( H \), \( a \in H_\alpha \) for every subgroup \( H_\alpha \) containing \( X \). Since each \( H_\alpha \) is a subgroup, and subgroups are closed under taking inverses, \( a^{-1} \in H_\alpha \) for every \( H_\alpha \). Therefore, \( a^{-1} \in H \). Thus, \( H \) is closed under taking inverses.

#### Conclusion:
Since \( H \) is non-empty, closed under the group operation, and closed under taking inverses, \( H \) satisfies the subgroup criteria. Therefore, \( H \) is a subgroup of \( G \).

This subgroup \( H \) is called the **subgroup of \( G \) generated by the subset \( X \)** because it is the smallest subgroup of \( G \) containing \( X \).

---

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the subgroup generated by \( X \) relate to the concept of a **normal subgroup**?
2. Can the subgroup generated by \( X \) be equal to \( G \)? Under what conditions?
3. If \( X \) is already a subgroup, how does \( H \) change?
4. How can you describe the **subgroup generated by multiple elements**?
5. What is the **free group** generated by a set, and how does it relate to this subgroup?

**Tip:** Remember that any subgroup must always contain the identity element, inverses, and be closed under the group operation. This is crucial when determining subgroups.

Let G be a group and X ⊂ G a subset (not necessarily a subgroup) of G. Let H be the intersection of all subgroups of G that contain X. Show that H is a subgroup of G. It is called the subgroup of G generated by the subset X.

Explore how to prove that the intersection of all subgroups of a group G containing a subset X forms a subgroup of G. This problem delves into group theory and the subgroup criterion, suitable for university-level students studying abstract algebra.

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