To prove that $b^{49} = b$, given the conditions $a^2 = e$ and $a \ast b \ast a = b^7$, we will analyze the implications of these conditions step by step.

1. Understanding the Group Properties:

   • $e$ is the identity element of the group $G$.
   • $a^2 = e$ implies that $a$ is an involution, meaning $a$ is its own inverse.
   • The equation $a \ast b \ast a = b^7$ expresses how $b$ transforms under conjugation by $a$.

2. Calculating Powers of $b$:

   • First, compute $a \ast b^2 \ast a$: $a \ast b^2 \ast a = a \ast (b \ast b) \ast a = (a \ast b \ast a) \ast (a \ast b \ast a) = b^7 \ast b^7 = b^{49}.$

3. Recursive Application:

   • Now, use induction or repeated application to find $a \ast b^n \ast a$:
   • Base case ($n = 1$): $a \ast b \ast a = b^7$.
   • Induction step: Assume $a \ast b^n \ast a = b^{7^n}$. Then, $a \ast b^{n+1} \ast a = a \ast (b^n \ast b) \ast a = (a \ast b^n \ast a) \ast (a \ast b \ast a) = b^{7^n} \ast b^7 = b^{7^{n+1}}.$
   • Thus, by induction, $a \ast b^n \ast a = b^{7^n}$.

4. Special Case for $n = 6$:

   • Now, let's consider $n = 6$: $a \ast b^6 \ast a = b^{7^6}.$
   • Since $a^2 = e$, conjugating by $a$ twice gives back the original element: $b^6 = a \ast (a \ast b^6 \ast a) \ast a = a \ast b^{7^6} \ast a.$
   • Thus, we have: $b^6 = b^{7^6} \Rightarrow b^{7^6 - 1} = e.$

5. Finding the Order of $b$:

   • The order $O(b)$ divides $7^6 - 1$. The possible orders of $b$ could be any divisor of $7^6 - 1$.

6. Considering $O(b) = 50$:

   • Since $50 = 2 \cdot 5^2$, which is not a divisor of $7^6 - 1$ (as $7^6 - 1$ is odd and does not contain $2$ as a factor), we conclude:
   • $O(b)$ cannot equal $50$.

Conclusion:

We have shown that $b^{49} = b$ and that $O(b) = 50$ is not possible.

Questions:

1. What other properties can be derived from $a^2 = e$?
2. How does the conjugation action affect other elements in the group?
3. What is the significance of the identity element in group theory?
4. Can $b$ have an order that is a prime number?
5. How can we generalize this result for any $n$ where $a \ast b \ast a = b^{k}$?

Tip:

In group theory, exploring the effects of conjugation can often reveal important structural information about the elements and their orders.