To prove that the inverse of the product of two elements $H$ and $K$ in a group $G$ is equal to the product of their inverses in reverse order, that is:

$(HK)^{-1} = K^{-1}H^{-1},$

we will use the properties of group operations and inverses.

Proof:

1. Group Inverse Definition:
   In a group $G$, for any element $g \in G$, its inverse $g^{-1}$ is defined as the element that satisfies: $g \cdot g^{-1} = g^{-1} \cdot g = e,$ where $e$ is the identity element of the group.

2. Goal:
   We need to show that: $(HK)^{-1} = K^{-1}H^{-1}.$ This means that when you multiply $HK$ by $K^{-1}H^{-1}$, the result should be the identity element $e$.

3. Left-Multiplication:
   Consider the product of $HK$ and $K^{-1}H^{-1}$: $(HK)(K^{-1}H^{-1}).$

   Using the associativity of group multiplication (since group operations are associative), we can rewrite this as: $H(K \cdot K^{-1})H^{-1}.$

   By the definition of an inverse, $K \cdot K^{-1} = e$, so this simplifies to: $H \cdot e \cdot H^{-1}.$

   Since $e$ is the identity element, multiplying by $e$ leaves the element unchanged, so we have: $H \cdot H^{-1}.$

   Again, by the definition of an inverse, $H \cdot H^{-1} = e$.

   Therefore: $(HK)(K^{-1}H^{-1}) = e.$

4. Right-Multiplication:
   Similarly, consider multiplying $K^{-1}H^{-1}$ by $HK$: $(K^{-1}H^{-1})(HK).$

   Again, using associativity, we rewrite this as: $K^{-1}(H^{-1}H)K.$

   Since $H^{-1}H = e$, this becomes: $K^{-1} \cdot e \cdot K.$

   As before, multiplying by $e$ leaves the element unchanged, so we have: $K^{-1} \cdot K = e.$

   Therefore: $(K^{-1}H^{-1})(HK) = e.$

Since both left and right multiplications result in the identity element $e$, it follows that: $(HK)^{-1} = K^{-1}H^{-1}.$

Conclusion:

We have proven that the inverse of the product of two elements in a group is the product of their inverses in reverse order. That is: $(HK)^{-1} = K^{-1}H^{-1}.$

---

Do you need more details on this proof or have any other questions?

Here are 5 related questions to expand on the concept:

1. What are some practical examples of groups where this property is used?
2. How do we extend this proof to the inverse of three or more elements?
3. Can this property be used to prove commutativity in certain types of groups?
4. How does the proof change if $H$ and $K$ are matrices?
5. How is this property used in abstract algebra applications?

Tip: Always verify group axioms (closure, associativity, identity, and inverses) when working with new groups to ensure that the elements follow the correct rules.