Let P be a subgroup of a group G and let Q = {xe G: xP = Px}. Is Q a subgroup of G?

Yes, Q is a subgroup of G.

To show this, we need to verify three conditions for Q to be a subgroup of G:

1. Q is non-empty: Since P is a subgroup of G, it contains the identity element e. Thus, eP = Pe = P, so e is in Q.
2. Q is closed under multiplication: Let x, y be in Q. Then we have xP = Px and yP = Py. We want to show that xy is in Q, i.e., (xy)P = P(xy). We have:

(xy)P = x(yP) (by associativity of multiplication) = x(Py) (since y is in Q) = (xP)y (since P is a subgroup of G) = Pyx (since x is in Q) = P(xy) (since y is in Q)

Thus, xy is in Q.

3. Q is closed under inverses: Let x be in Q. We want to show that x^-1 is in Q, i.e., (x^-1)P = P(x^-1). We have:

(x^-1)P = (Px)^-1 (since P is a subgroup of G) = (xP)^-1 (since x is in Q) = P(x^-1) (since P is a subgroup of G)

Thus, x^-1 is in Q.

Since Q satisfies all three conditions for being a subgroup of G, we conclude that Q is a subgroup of G.