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CM121A, Abstract Algebra Solution Sheet 5
Solutions to questions 5-7 from exercise sheet 4, and questions
1-4
from exercise sheet 5
Sheet 4:
5. Consider the following elements of S5:
=
1 2 3 4 52 3 1 5 4
, =
1 2 3 4 55 4 3 2 1
.
(Recall this means : {1, 2, 3, 4, 5} {1, 2, 3, 4, 5} is defined
by (1) = 2,(2) = 3, (3) = 1, etc.)
(a) Express and using cycle notation.Solution: = (123)(45), =
(15)(24).
(b) Compute .Solution: Since ((1)) = (5) = 4, ((2)) = (4) = 5,
((3)) =(3) = 1, ((4)) = (2) = 3 and ((5)) = (1) = 2, we have:
=
1 2 3 4 54 5 1 3 2
,
or in cycle notation = (143)(25).
(c) Compute the inverse of in S5.Solution: We have 1(2) = 1,
1(3) = 2, 1(1) = 3, 1(5) = 4and 1(4) = 5, so:
1
= 1 2 3 4 5
3 1 2 5 4
,
or in cycle notation this is (132)(45).
6. Suppose that A, B and C are sets, and that f : A B and g : B
Care functions. Prove the following:
(a) Ifg f is injective, then f is injective.Solution: If f(a) =
f(a), then g(f(a)) = g(f(a)), which meansthat (g f)(a) = (g f)(a).
If g f is injective, this implies thata = a, so f is also
injective.
(b) Ifg f is surjective, then g is surjective.Solution: Suppose
that c C. If (g f) is surjective, then (g f)(a) = c for some a A.
This means that g(f(a)) = c, so g(b) = cfor some b B, namely b =
f(a). Therefore g is surjective.
7. Suppose that A and B are sets, and that f : A B and g : B A
arefunctions. We say that g is a left inverse of f if g f = idA.
Similarly gis a right inverse of f if f g = idB.
(a) Prove that f has a right inverse (i.e., f g = idB for some
functiong : B A) if and only if f is surjective.Solution: Suppose
that f has a right inverse g, so f g = idB. SinceidB is surjective,
part b) of the preceding problem implies that f issurjective as
well.Suppose that f is surjective. This means that for each b B,
thereis some a A such that f(a) = b. Define a function g : B A
by choosing such an a for each b. Then f(g(b)) = f(a) = b, sof g
= idB, and so f has a right inverse.
(b) Suppose that A is not the empty set. Prove that f has a left
inverseif and only if f is injective.Solution: Suppose that f has a
left inverse g, so g f = idA. Then
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g f is injective, and therefore so is f by part a) of the
precedingproblem.Suppose that f is injective. Define a function g :
B A as follows:Suppose that b B. Ifb = f(a) for some a A, then
there is exactlyone such a (since f is injective) and we define
g(b) to be this a. Westill need to define g(b) if b is not in the
range of f. Since A = ,there is some element a0 A, and we define
g(b) to be a0 for all suchb. Then g(f(a)) = a for all a A, because
ifb = f(a), then g(b) = a.Note: We had to assume A was non-empty
for the following reason:IfA = and B is non-empty, then there is a
function from f : A B.We just have to define f(a) for every a A,
but there are no elementsofA, so we dont have to do anything to
define f. And in fact theresonly one function from the empty set to
B, and it is injective. Onthe other hand there are no functions g :
B A (since there are nopossible values), so even though f is
injective, it cant possibly havea left inverse.
Sheet 5:
1. Suppose that G is a group with identity element e, and that g
and h areelements of G. Prove that there is a unique element x G
such thatg(xh) = e.Solution: Let x = g1h1. Then
g(xh) = g((g1h1)h)= g(g1(h1h)) (associativity)= g(g1e) (since
h1h = e)= (gg1)e (associativity)= gg1 (definition of identity)
= e.
This is the only solution since if g(xh) = e, then
x = exe = (g1g)x(hh1) = g1(gxh)h1 = g1eh1 = g1h1.
(again using associativity, definition of inverses and
identity).Remark: At this stage you should be comfortable enough
with the con-sequences of associativity to omit the parentheses as
Ive just done (forexample, writing x = exe instead of x = ex =
(ex)e, etc.)
2. Suppose that G and g is an element of G such that g2 = g.
Prove thatg = e (the identity element of G).Solution: Since gg = g
= ge, the cancellation law implies that g = e.
3. Suppose that G is a group such that g2 = e (the identity
element) for allg G. Prove that G is abelian.Solution: Suppose that
g, h G. Applying the equation to the elementgh G gives ghgh = (gh)2
= e. Multiplying the equation by g on the leftand h on the right
(and applying associativity) gives gives g(ghgh)h =geh = gh. Since
gg = hh = e, it follows (again using associativity) thatg(ghgh)h =
(gg)(hg)(hh) = e(hg)e = hg, and therefore gh = hg. Wevenow shown
that gh = hg for all g, h G, so G is abelian.
4. Suppose that G is a group and g G. Prove that (gm)n = gmn for
allm, n Z as follows:
(a) First prove the formula holds for all m Z and n N. (Use
inductionon n and the fact proved in lecture that gagb = ga+b for
all a, b Z.)
Solution: For n = 1, the formula just says gm = gm. Supposenow
that n > 1 and that (gm)n1 = gm(n1). Then the formulagagb = ga+b
gives
(gm)n = (gm)n1gm = gm(n1)gm = gm(n1)+m = gmn.
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(b) Deduce that the formula holds for all m, n Z.Solution: For n
= 0, we have (gm)0 = e = g0 = gmn. Suppose now
that n < 0. Then (gm)n is defined as ((gm)n)1
(i.e., the inverseof (gm)n). Since n > 0, weve already shown
that (gm)n =gm(n) = gmn, so we only need to show that gmn is the
inverse ofgmn, but this follows from the formula
gmngmn = gmn+(mn) = g0 = e
(or even from the definitions: ifmn = 0, then these are both
definedas e; otherwise, the one with the negative exponent is
defined as theinverse of the other).
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