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.