If x and y are elements in a group G and if x5 = y3 = e, where e is the identity of G, then the inverse of x2yx4y2 must be

  1. y2xy2x4
  2. yxy2x3
  3. yx6y6x3
  4. x4y2x2y

Answer (Detailed Solution Below)

Option 2 : yxy2x3

Detailed Solution

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The correct answer is: Option 2

Key Points

Concept:

The inverse of a product of elements in a group is the reverse product of their inverses. Given \( x^5 = e \) and \( y^3 = e \), we use:
1. \( x^{-1} = x^4 \) (since \( x \cdot x^4 = e \)).
2. \( y^{-1} = y^2 \) (since \( y \cdot y^2 = e \)).

Given:

Element to invert: \( x^2 y x^4 y^2 \)
Conditions: \( x^5 = y^3 = e \) (identity).

Steps:

1. Apply the inverse property \( (ab)^{-1} = b^{-1}a^{-1} \) in reverse order:
\( \left( x^2 y x^4 y^2 \right)^{-1} = (y^2)^{-1} (x^4)^{-1} y^{-1} (x^2)^{-1} \).

2. Compute each inverse:
- \( (y^2)^{-1} = y \) (since \( y^3 = e \implies y^4 = y \)).
- \( (x^4)^{-1} = x \) (since \( x^4 \cdot x = e \)).
- \( y^{-1} = y^2 \).
- \( (x^2)^{-1} = x^3 \) (since \( x^2 \cdot x^3 = e \)).

3. Substitute and simplify:
\( y \cdot x \cdot y^2 \cdot x^3 = y x y^2 x^3 \).

Verification:
Multiply \( x^2 y x^4 y^2 \) by \( y x y^2 x^3 \):
- \( y^2 \cdot y = e \), \( x^4 \cdot x = e \), \( y \cdot y^2 = e \), \( x^2 \cdot x^3 = e \).
Result: \( e \) (identity), confirming correctness.

Final Answer:Option 2 (\( y x y^2 x^3 \))

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