Math 3124 Friday, September 12
Second Homework Solutions
1. Problem 5.14 on page 34 Let H={f:R→R|f(x)6=0 for all x∈R}. For
f,g∈H, define f g by (f g)(x) = f(x)g(x)for all x∈R. Then f g ∈H. Verify that with
this operation His a group. How does this group differ from the group of invertible
mappings in M(R)? Also, is this group Habelian?
First note that we do indeed have a binary operation, because if f,g∈H, then f g:R→
R, and f g(x) = f(x)g(x)6=0 for all x∈R. We now have the three axioms for a group
to check.
(a) Associativity. If f,g,h∈Hand x∈R, then
(f g)h(x)=(fg)(x)h(x) = f(x)g(x)h(x) = f(x)(gh)(x) = f(gh)(x)
for all x∈R. Therefore (f g)h=f(gh)which proves associativity.
(b) Identity. This is the function ι:R→Rdefined by ι(x) = 1 for all x∈R.
(c) Inverses. The inverse of the function fis the function 1/f, where (1/f)(x) =
1/f(x)(note that this is well defined because f(x)6=0 for all x∈R).
We have now verified that His a group. It differs in many ways from the group of
invertible mappings in M(R). For one thing His abelian, because if f,g∈H, then
(f g)(x) = f(x)g(x) = g(x)f(x) = (g f )(x)for all x∈Rand hence f g =g f , while the
group of invertible mappings in M(R)is nonabelian.
2. Let A,B∈G. Then certainly AB is a 2 ×2 matrix with real entries. Furthermore
(AB)(AB)0=A(BB0)A0=AIA0=AA0=I
and we see that AB ∈G, so we have closure and indeed matrix multiplication is an
operation on G. Also matrix multiplication is associative, and the identity is I(note
that I∈Gbecause II0=II =I). Finally if A∈G, then Ais an invertible matrix with
inverse A0and we have A0A=I. Therefore
I=I0= (A0A)0=A0(A0)0
which shows that A0∈Gand it follows that all elements have inverses. This completes
the proof that Gis a group.
3. Let
α=1 2 3 4
3 4 1 2and β=1 2 3 4
4 3 1 2.
(a) β◦α. This sends 1 →3→1, 2 →4→2, 3 →1→4, 4 →2→3. Therefore β◦α
is 1 2 3 4
1 2 4 3
