With that misunderstanding cleared up, the proof seems pretty straightforward:-
Consider f in Auto(G). f is surjective and injective as well as homomorphic. The inverse f^-1 is surjective as well. Given any x in G, x = f^-1(f(x)) such that x is in the image of f^-1 since the image of f(x)...
f is a homomorphism since it is part of the automorphism group of G. So therefore:-
e' = f(e) = f(x x^-1) = f(x)f(x^-1) .. where e' is the identity in the image of f. But this means that f(x^-1) is f^-1(x) since we can also have e' = f(x^-1 x) = f(x^-1)f(x) by similar logic.
The problem is as follows:-
Statement of the problem: Given group G, show that the automorphism group of G is a subgroup of the permutation group of G.
I can show that Auto(G) is a subset of Perm(G) easily. So I have to show that subgroup conditions hold: (1) for each x in Auto(G), x inverse...
For part 1, you can write a piecewise function but it is just typing out what maximum means.
For the second part, you can't necessarily recover A from C. Suppose that B>A. Then A could be any value less than C.
Your other condition is that its monotonic increasing in n. And there's a theorem that says it converges to the sup of {arctan(2n): n>0}, which is pi/2.
GRE Math subject exam Prep-- Pre-calc
For those of you who are preping, or have taken it already, what did you use to prep for the pre-calc section of the GRE subject exam in math?