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Let's solve question 2a from the image:

### Question 2a:
In a group \( (G, *) \), prove that \( (a * b)^{-1} = b^{-1} * a^{-1} \) for all \( a, b \in G \).

### Solution:

We are asked to prove that for any elements \( a \) and \( b \) in a group \( (G, *) \), the inverse of the product of \( a \) and \( b \) is the product of the inverses of \( b \) and \( a \), but in reverse order.

#### Proof:

Let \( a, b \in G \). By the definition of the inverse in a group, we know that for any element \( x \in G \), \( x * x^{-1} = x^{-1} * x = e \), where \( e \) is the identity element in the group.

We want to show that:
\[
(a * b)^{-1} = b^{-1} * a^{-1}
\]

**Step 1:** We know that \( (a * b)^{-1} \) is the element that, when multiplied by \( a * b \), gives the identity element \( e \). So we need:
\[
(a * b) * (a * b)^{-1} = e
\]

**Step 2:** Now assume \( (a * b)^{-1} = b^{-1} * a^{-1} \), and check if this satisfies the group inverse property. We check if multiplying \( a * b \) by \( b^{-1} * a^{-1} \) gives the identity element:
\[
(a * b) * (b^{-1} * a^{-1})
\]

Using the associativity property of the group operation, we can rewrite this as:
\[
= a * (b * b^{-1}) * a^{-1}
\]
Since \( b * b^{-1} = e \) (by the definition of the inverse), this simplifies to:
\[
= a * e * a^{-1} = a * a^{-1}
\]
Again, by the definition of the inverse, \( a * a^{-1} = e \). Thus, we have:
\[
(a * b) * (b^{-1} * a^{-1}) = e
\]

Since \( (a * b) * (b^{-1} * a^{-1}) = e \), it follows that \( (a * b)^{-1} = b^{-1} * a^{-1} \), as required.

#### Conclusion:
We have proved that \( (a * b)^{-1} = b^{-1} * a^{-1} \) for all \( a, b \in G \).

---

Would you like further clarification on any step or more details on group theory concepts? Here are five related questions to explore:

1. What is the identity element in a group, and why is it important for proving inverse properties?
2. Can you prove the inverse of a product in non-abelian groups using a different method?
3. How does the concept of associativity play a role in the proof of \( (a * b)^{-1} = b^{-1} * a^{-1} \)?
4. What is the difference between abelian and non-abelian groups in terms of product inverses?
5. Can you give an example of a group where \( (a * b)^{-1} \neq a^{-1} * b^{-1} \) and explain why?

**Tip:** Always remember that in a group, the operation is not necessarily commutative (i.e., \( a * b \neq b * a \)), so careful attention is needed to the order of elements when proving properties like this one.

In a group (G, *), prove that (a * b)^(-1) = b^(-1) * a^(-1) for all a, b in G.

Explore how to prove that (a * b)^(-1) = b^(-1) * a^(-1) in a group (G, *). This detailed proof uses core group theory concepts such as the inverse element property and associativity. Suitable for undergraduate students studying abstract algebra.

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