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.