To determine if $H$ is a subgroup of $G$, we need to check the subgroup criteria, which are:

1. Closure: If $a$ and $b$ are elements of $H$, then $ab$ must also be an element of $H$.
2. Identity element: The identity element $e$ of the group $G$ must be in $H$.
3. Inverses: For every element $a \in H$, its inverse $a^{-1}$ must also be in $H$.

The condition given is:

• If $a$ and $b$ are in $H$, then $(a^{-1})(b^{-1})$ is also in $H$.

This condition suggests that $H$ might have the necessary properties for closure under the operation and inverses, but we still need to examine whether it guarantees the subgroup criteria.

1. Closure:

The condition provided, $(a^{-1})(b^{-1}) \in H$ for $a, b \in H$, does not directly tell us whether $ab \in H$, which is required for closure. Therefore, the closure property is not guaranteed by this condition.

2. Identity element:

To check if the identity element $e$ is in $H$, consider that for any $a \in H$, $a^{-1}$ must also be in $H$ (since $(a^{-1})(a^{-1}) \in H$ by the condition). This suggests that the identity element might be in $H$, because $e^{-1} = e$, so we can expect $e$ to be in $H$ by virtue of the inverse condition.

3. Inverses:

If $a \in H$, the condition ensures that $a^{-1} \in H$, since $a^{-1}$ will appear as part of the product $(a^{-1})(a^{-1})$, which is in $H$.

Conclusion:

Although the given condition ensures that inverses are in $H$ and possibly the identity element is in $H$, it does not guarantee closure (i.e., $ab \in H$ for $a, b \in H$). Thus, $H$ is not necessarily a subgroup of $G$ based on this condition alone.

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Would you like more details or further exploration into any part of this? Here are some related questions to expand your understanding:

1. What does the closure property of a subgroup specifically require in terms of elements of the group?
2. How does the identity element relate to the inverse of an element in a subgroup?
3. What is the significance of the given condition $(a^{-1})(b^{-1}) \in H$ in group theory?
4. Can you construct an example where the given condition holds but $H$ is not a subgroup?
5. How would you prove closure in a subgroup more directly?

Tip: Always ensure the three subgroup conditions (closure, identity, and inverses) are verified when checking if a subset is a subgroup!