Advanced High-School Mathematics

238 CHAPTER 4 Abstract Algebra

that f is also one-to-one. (Is this obvious?) Finally, let xk, xl ∈ G;
ifk+l ≤n−1, thenf(xkxl) =f(xk+l) = yk+l =ykyl =f(xk)f(xl).
However, ifk+l ≥n, then we need to dividen intok+l and get a
remainderr, where 0≤r≤n−1, sayk+l=qn+r, whereq is the
quotient andris the remainder. We therefore have that

f(xkxl) = f(xk+l)

= f(xqn+r)

= f(xqnxr)

= f(eGxr) (sincexqn=eG, the identity ofG)

= f(xr)

= yr (by definition off)

= eHyr (whereeH is the identity ofH)

= yqnyr (sinceyqn=eH)

= yqn+r

= yk+l

= ykyl.

Exercises

1. Letf :G 1 →G 2 be a homomorphism of groups, and letH 1 ⊆G 1
   be a subgroup of G 1. Prove that the image, f(H 1 ) ⊆ G 2 is a
   subgroup ofG 2.


2. (A little harder) Letf :G 1 →G 2 be a homomorphism of groups,
   and let H 2 ⊆ G 2 be a subgroup of G 2. Set f−^1 (H 2 ) = {g 1 ∈
   G 1 |f(g 1 )∈H 2 }. Prove thatf−^1 (H 2 ) is a subgroup ofG 1.


3. Let GL 2 (R) be the group of 2×2 matrices with real coefficients
   and determinant not 0, and letSbe a nonsingular matrix. Define
   the mappingf : GL 2 (R)→GL 2 (R) by setting f(A) = SAS−^1.
   Prove thatf is an isomorphism of GL 2 (R) onto itself.


4. Again, let GL 2 (R) be the group of 2×2 matrices with real coeffi-
   cients and determinant not 0. Show that the determinant defines