To prove that the inverse of any element of a subgroup is the same as the inverse of the same element regarded as an element of the group, let's proceed step by step:

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Definitions and Setup:

1. Let $G$ be a group, and $H$ be a subgroup of $G$, which means:

   • $H \subseteq G$.
   • $H$ is closed under the group operation in $G$.
   • $H$ contains the identity element of $G$, say $e$.
   • $H$ is closed under inverses.

2. Let $h \in H$, and $h^{-1}$ be the inverse of $h$ in $H$. We need to show that $h^{-1}$, as defined in $H$, is the same as the inverse of $h$ in $G$.

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Proof:

1. $h^{-1}$ in $H$ satisfies the inverse property in $G$:

• Since $H$ is a subgroup of $G$, $h \in H \implies h \in G$ (by definition of a subgroup).
• The inverse $h^{-1}$ in $H$ satisfies: h \cdot h^{-1} = e_H \quad \text{(the identity in H)}. But $H$ is a subgroup, and the identity $e_H$ of $H$ is the same as the identity $e_G$ of $G$ (by definition of subgroup). Thus: $h \cdot h^{-1} = e_G.$
• Hence, $h^{-1}$ satisfies the inverse property for $h$ in $G$.

2. Uniqueness of inverses in $G$:

• In a group $G$, the inverse of any element is unique.
• Since $h^{-1}$ satisfies the defining property of the inverse in $G$ (i.e., $h \cdot h^{-1} = e_G$), it must be the same as the inverse of $h$ when regarded as an element of $G$.

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Conclusion:

Thus, the inverse $h^{-1}$ of an element $h$ in $H$ is the same as the inverse of $h$ when regarded as an element of $G$.

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Follow-up Questions:

1. What are the defining properties of a subgroup?
2. How does the uniqueness of inverses in a group help in proving subgroup properties?
3. Can you provide examples where this property holds for specific subgroups of well-known groups?
4. How does the closure of a subgroup under the group operation assist in this proof?
5. What are the implications of the identity element being the same in a group and its subgroup?

Tip:

When dealing with subgroups, always remember that the identity and inverses are consistent with those in the parent group,