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```§3.1 Solutions
36. For each binary operation ∗ defined on a set below, determine whether or not ∗ give a group structure on the
set. If it is not a group, say which axioms fail to hold.
We evidently get to assume that each of the operations is a binary operation, so we don’t need to check the closure
property. Note that if there is no identity element, you can’t even ask the question about existence of inverses.
1
1
+ .
x y
The operation is not associative since (1 ∗ 2) ∗ 3 = 1, but 1 ∗ (2 ∗ 3) = 11/5.
Suppose that e is an identity element. Then e ∗ 1 = 1, and this forces 1/e = 0, a contradiction.
(c) Define ∗ on Q+ by x ∗ y =
(d) Define ∗ on R by x ∗ y = x + y − 1.
To show that the operation is associative, we have the following calculations for all a, b, c ∈ G.
a ∗ (b ∗ c) = a ∗ (b + c − 1) = a + (b + c − 1) − 1 = a + b + c − 2
(a ∗ b) ∗ c = (a + b − 1) ∗ c = (a + b − 1) + c − 1 = a + b + c − 2
To find an identity element we need to solve a ∗ e = a (this is enough since the operation is symmetric in x
and y and therefore commutative). We get a + e − 1 = a, so e = 1 and is independent of a. Checking this gives
a ∗ 1 = a + 1 − 1 = a for all a ∈ R.
Given a ∈ G, to find its inverse we must solve a ∗ x = 1, or a + x − 1 = 1, so we get x = 2 − a. This does work
because a ∗ (2 − a) = a + (2 − a) − 1 = 1.
(e) Define ∗ on R× by x ∗ y = xy + 1.
The operation is not associative since (1 ∗ 2) ∗ 3 = 10 but 1 ∗ (2 ∗ 3) = 8.
Suppose that e is and identity element. Then 1 ∗ e = 1, so e + 1 = 1 and therefore e = 0. But then e fails to work
as an identity for 2, since 2 ∗ e = 2 · 0 + 1 = 1 6= 2.
(g) Define ∗ on R2 by (x1 , y1 ) ∗ (x2 , y2 ) = (x1 x2 , y1 x2 + y2 ).
To check the associative law, we have the following calculations.
((a1 , b1 )∗(a2 , b2 ))∗(a3 , b3 ) = (a1 a2 , b1 a2 +b2 )∗(a3 , b3 ) = ((a1 a2 )a3 , (b1 a2 +b2 )a3 +b3 ) = (a1 a2 a3 , b1 a2 a3 +b2 a3 +b3 )
(a1 , b1 ) ∗ ((a2 , b2 ) ∗ (a3 , b3 )) = (a1 , b1 ) ∗ (a2 a3 , b2 a3 + b3 ) = ((a1 (a2 a3 ), b1 (a2 a3 ) + b2 a3 + b3 )
To check for an identity, we need to solve (x, y) ∗ (e, f ) = (x, y). Thus (xe, ye + f ) = (x, y), so xe = x and
ye + f = y. The solution is e = 1 and f = 0. To check this solution, we have (a, b) ∗ (1, 0) = (a · 1, b · 1 + 0) = (a, b)
and (1, 0) ∗ (a, b) = (1 · a, 0 · a + b) = (a, b).
Finally, to find the inverse of (a, b) we must solve the equation (a, b) ∗ (x, y) = (1, 0), or (ax, bx + y) = (1, 0). This
forces x = 1/a, and now we are in trouble because the problem does not assume that a 6= 0. We conclude that no
element of the form (0, b) has a multiplicative inverse. (Note: if a 6= 0, then the inverse of (a, b) is (1/a, −b/a).)
x y x, y ∈ R .
(h) Use matrix multiplication to define ∗ on
0 0 Matrix multiplication is always associative.
e f
x y
x y
=
or xe = x and xf = y. The solution for f
To find an identity we need to solve
0 0
0 0
0 0
is not independent of x and y, so there is no identity element in this form.
37. Let G be a group, and suppose that a, b, c ∈ G. Solve the equation axc = b.
First suppose that the equation does have a solution x ∈ G. We multiply both sides of the equation (on the left)
by a−1 to get xc = a−1 b, and then multiply both sides of the equation (on the right) by c−1 to get x = a−1 bc−1 .
This actually proves the uniqueness of the solution–if there is a solution, it must be x = a−1 bc−1 . To check that our
calculation actuallyl produces a solution, we substitute for x, and get axc = a(a−1 bc−1 )c = (aa−1 )b(c−1 c) = ebe = b.
Note: Even in high school algebra you need to be careful about what you are actually doing. Sometimes “solving” the
equation may produce extraneous roots, so you must substitute back into the equation to check that you have really
found a solution.
39. Let G be a group with operation · and let a ∈ G. Define a new operation ∗ on the set G by x ∗ y = x · a · y, for
all x, y ∈ G. Show that G is a group under the operation ∗.
It is clear that x ∗ y is a unique element of G, and so ∗ is a binary operation.
To check associativity, we have the following calculations for all x, y, z ∈ G.
(x ∗ y) ∗ z = (xay) ∗ z = xayaz
x ∗ (y ∗ z) = x ∗ (yaz) = xayaz
To find an identity element, for g ∈ G we need to solve x ∗ g = g and g ∗ x = g. Then we must have xag = g, so
x = a−1 . You can check that a−1 is both a left and a right identity element.
To find the inverse of g, we must solve g ∗ x = a−1 and x ∗ g = a−1 . Thus gax = a−1 and xag = a−1 . Both
equations have the solution x = (aga)−1 , and then you can check that the inverse of g in the new group is (aga)−1 .
40. Let G be a group, with operation ·. Define a new operation by x ∗ y = y, for x, y ∈ G. Is G a group under this
operation? If not, which axioms hold and which do not?
The given multiplication defines a function from G × G into G, so we have a binary operation.
The operation is associative, since (x ∗ y) ∗ z = y ∗ z = z and x ∗ (y ∗ z) = x ∗ z = z.
To find an identity element, ‘ we can solve x ∗ y = y by letting x = e, the original identity element of G. But on
the other side, y ∗ x = y is impossible to solve if y has more than one element, since y ∗ x = x. Thus G has a left
identity but not a right identity.
Note: a similar argument shows that G has left inverses for each element, but not right inverses.
44. Let T be the set of functions tc,d : R2 → R2 defined by tc,d (x1 , x2 ) = (x1 + c, x2 + d), where c, d are any elements
of R. Show that T is a group under composition of functions.
You can check that tc,d ◦ ta,b = ta+c,b+d , and so the closure property holds. The associative law holds since the
operation is composition of functions. From the formula, the identity element is t0,0 , and the inverse function for tc,d
is t−c,−d .
45. Let G be a group. For x, y ∈ G, define x ∼ y if there exists some element a ∈ G such that y = axa−1 . Show
that ∼ defines an equivalence relation on G. Note: This is a general case of Exercise 2.3.15 in the text.
Reflexive property: x ∼ x using either a = e or a = x.
Symmetric property: if x ∼ y, then there exists a ∈ G with y = axa−1 , so x = a−1 y(a−1 )−1 , and a−1 ∈ G.
Transitive property: if x ∼ y and y ∼ z, then there exist a, b ∈ G with y = axa−1 and z = byb−1 . Then ba ∈ G
and x ∼ z since z = b(axa−1 )b−1 = (ba)x(ba)−1 .
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