Question

In a group \(G\), let \(x^5=e\) (\(e\): identity element of \(G\)) and \(xyx^{-1}=y^2\) for \(x,y\in G\). If \(y\ne e\), then the order of \(y\) is ____.

• A. 25
• B. 30
• C. 31
• D. 32

Hint:

Conjugate y by x five times to get y equals y to the power 32, then use that 31 is prime.

Answer 1

The Correct Option is C

Step 1: Apply the conjugation relation repeatedly.
We are given \(xyx^{-1}=y^2\). Conjugating again by \(x\): \[x^2yx^{-2}=x(xyx^{-1})x^{-1}=xy^2x^{-1}=(xyx^{-1})^2=y^4.\]

Step 2: Continue the pattern.
By induction, \(x^kyx^{-k}=y^{2^k}\) for every \(k\ge 1\). Indeed \(x^3yx^{-3}=x(x^2yx^{-2})x^{-1}=xy^4x^{-1}=(xyx^{-1})^4=y^8\), similarly \(x^4yx^{-4}=y^{16}\), and \(x^5yx^{-5}=y^{32}\).

Step 3: Use \(x^5=e\).
Since \(x^5=e\), the left side becomes \(x^5yx^{-5}=eye^{-1}=y\). So \[y=y^{32}\implies y^{31}=e.\]

Step 4: Determine the order of \(y\).
The order of \(y\) must divide 31. Since 31 is prime, the order of \(y\) is either 1 or 31. Order 1 would mean \(y=e\), which is excluded. Hence the order of \(y\) is 31.

Conclusion: \[\boxed{\text{Order of } y = 31}\]