Question about Sylow Theorem? (modern algebra 1)

Show that if P is a Sylow p-subgroup of G and g belongs to G then the conjugate map x-> g^-1 x g is a bijection and thus |g^-1 P gI = |P|

I'll call the conjugate map F: P --> g^(-1) P g.

(i) F is 1-1.
Suppose that F(x) = F(y) for some x, y in P.
Then, g^(-1) x g = g^(-1) y g
==> xg = yg, via left cancellation
==> x = y, via right cancellation.

(ii) F is onto:
Let y be in g^(-1) P g.
So, y = g^(-1) xg for some x in P.
Then, F(x) = y, as required.
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By (i) and (ii), F is a bijection, and thus |P| = |g^(-1) P g|.

I hope this helps!