 Let (𝐺,∗) be a group and 𝑎, 𝑏 ∈ 𝐺 such that 𝑎
2 = 𝑒 and 𝑎 ∗ 𝑏 ∗ 𝑎 = 𝑏
7
.prove 
that 𝑏
49 = 𝑏. Can 𝑂(𝑏) = 50? 

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.

Let (𝐺,∗) be a group and 𝑎, 𝑏 ∈ 𝐺 such that 𝑎^2 = 𝑒 and 𝑎 ∗ 𝑏 ∗ 𝑎 = 𝑏^7. Prove that 𝑏^{49} = 𝑏. Can 𝑂(𝑏) = 50?

Explore the proof that b^49 = b in a group (G,*) under the conditions a^2 = e and a * b * a = b^7. The solution delves into group properties, the order of elements, and shows why O(b) cannot equal 50.

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