let g be an abelian group under multiplitication and H = x^2 ;x belongs  to g , is h subgrouo of g

Let \( G \) be an Abelian group under multiplication, and \( H = \{x^2 \mid x \in G\} \). We want to determine whether \( H \) is a subgroup of \( G \).

To show that \( H \) is a subgroup of \( G \), it must satisfy the following criteria:

1. **Closure**: For any \( h_1, h_2 \in H \), the product \( h_1 h_2 \in H \).
2. **Identity element**: The identity element of \( G \) (denoted by \( e \)) must be in \( H \).
3. **Inverses**: For every element \( h \in H \), its inverse \( h^{-1} \) must also be in \( H \).

### Step 1: Closure
Let \( h_1 = x^2 \) and \( h_2 = y^2 \) be elements of \( H \), where \( x, y \in G \). Since \( G \) is Abelian, we have:

\[
h_1 h_2 = x^2 y^2 = (xy)^2
\]

Since \( xy \in G \) (because \( G \) is closed under multiplication), we see that \( (xy)^2 \in H \). Therefore, \( H \) is closed under multiplication.

### Step 2: Identity Element
The identity element of \( G \) is \( e \), which satisfies \( e^2 = e \). Therefore, \( e \in H \). So, \( H \) contains the identity element.

### Step 3: Inverses
Let \( h = x^2 \in H \). We need to check if \( h^{-1} \in H \). The inverse of \( h \) in \( G \) is \( (x^2)^{-1} = (x^{-1})^2 \). Since \( x^{-1} \in G \) (because \( G \) is a group), we have \( (x^{-1})^2 \in H \). Hence, \( H \) is closed under inverses.

### Conclusion
Since \( H \) satisfies closure, contains the identity element, and is closed under inverses, we can conclude that \( H \) is indeed a subgroup of \( G \).

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Would you like further details on any of the steps or have any other questions?

Here are 5 related questions to expand this concept:
1. What is the significance of the Abelian property in this proof?
2. Can you provide an example of a non-Abelian group where this subgroup property might fail?
3. How does the concept of normal subgroups relate to this scenario?
4. What happens if we consider \( H = \{x^n \mid x \in G \} \) for some \( n > 2 \)?
5. Could \( H \) be a proper subgroup of \( G \) in all cases?

**Tip**: Always check the Abelian property when dealing with subgroup proofs, as it often simplifies the multiplication operations.

Let G be an abelian group under multiplication and H = {x^2 | x belongs to G}, is H a subgroup of G?

Explore whether the set H = {x^2 | x belongs to G} forms a subgroup of an abelian group G under multiplication. This problem applies key group theory concepts such as closure, identity element, and inverses. Suitable for undergraduate-level students studying abstract algebra.

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