We are given a group $G$ and two elements $a, b \in G$, and we are asked to prove that:

$(a * b)' = a' * b' \quad \text{if and only if} \quad a * b = b * a$

Where the apostrophe ($'$) represents the inverse of an element, and the operation $*$ denotes the group operation.

Step 1: Proving the "if" direction (i.e., if $a * b = b * a$, then $(a * b)' = a' * b'$).

Assume that $a * b = b * a$ (i.e., $a$ and $b$ commute).

To prove: $(a * b)' = a' * b'$.

1. Start by using the property of the inverse element in a group: $(a * b) * (a * b)' = e$ where $e$ is the identity element of the group.

2. Since $a * b = b * a$, we can substitute $b * a$ for $a * b$ in the equation: $(a * b) * (a * b)' = (b * a) * (a * b)' = e$

3. By associativity, we can group the terms: $b * (a * (a * b)') = e$

4. Notice that $a * (a * b)'$ should be the identity element, as the inverse of an element is defined to cancel with the element itself: $a * (a * b)' = b'$

5. Therefore, the equation simplifies to: $b * b' = e$ which is true because $b'$ is the inverse of $b$.

Hence, $(a * b)' = a' * b'$.

Step 2: Proving the "only if" direction (i.e., if $(a * b)' = a' * b'$, then $a * b = b * a$).

Assume that $(a * b)' = a' * b'$.

To prove: $a * b = b * a$.

1. Start by using the fact that $(a * b) * (a * b)' = e$ and substitute the given condition $(a * b)' = a' * b'$: $(a * b) * (a' * b') = e$

2. Using the associativity of the group, we can write: $a * (b * (a' * b')) = e$

3. Rearrange the terms, applying the property of inverses: $a * ((b * a') * b') = e$

4. Now, simplify $(b * a') * b'$. Since $a'$ is the inverse of $a$, we know: $(b * a') * b' = b * (a' * b') = b * e = b$

5. So, the equation becomes: $a * b = e$ which proves that $a * b = b * a$, because we are operating in a group.

Thus, we have shown that if $(a * b)' = a' * b'$, then $a * b = b * a$.

Conclusion:

We have shown both directions of the statement:

• If $a * b = b * a$, then $(a * b)' = a' * b'$.
• If $(a * b)' = a' * b'$, then $a * b = b * a$.

Thus, the given equivalence holds.

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Would you like further details or explanations on any part of this proof?

Here are 5 related questions for further exploration:

1. How does the commutativity of the group operation affect the structure of the group?
2. What happens in the proof if the group is non-abelian?
3. How can we apply this result to specific groups, such as $\mathbb{Z}_n$ or $\mathbb{R}$?
4. What other group properties could be used to prove similar results involving inverses?
5. What is the significance of the identity element in group theory proofs?

Tip: In group theory, always check if the group operation is commutative (abelian) before proceeding with simplifications, as this often simplifies proofs dramatically.