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(Solved) Math 417, Spring 2017. Homework 1 Solutions Due: Thursday, January 26, 2016 (1) Exercise 1 on page 12 of the notes Solution. A binary operation is a


Can you help me on the topic this week? I think it is on  Homomorphisms and Isomorphisms. Attached the materials (HW3 and updated notes and the given answer for last two HWs just for references ).

Math 417, Spring 2017. Homework 1 Solutions
Due: Thursday, January 26, 2016
(1) Exercise 1 on page 12 of the notes
Solution. A binary operation is a map µ : X × X −→ X and so
can be pictured as a table:
µ x y
µ ←→ x ? ?
y ? ?
there are no restrictions on how we fill in the table, and we have
two choices for each spot, so there are a total of 24 = 16 binary
operations.
(i) commutative operations: now we need to impose that µ(x, y) =
µ(y, x) so we have three spots to choose, and two choices for each
spot, so a total of 23 = 8 commutative binary operations.
(iii) associative operations: now we need to impose µ(µ(a, b), c) =
µ(a, µ(b, c)). In the interest of space, we will use multiplicative notation µ(x, y) = xy.
From (xx)x = x(xx) we see that either xx = x or xy = yx. Similarly
(yy)y = y(yy) tells us that either yy = y or xy = yx.
Thus if we first look for non-commutative operations, we know that
xx = x, yy = y, xy 6= yx leaving us with two possibilities,
µ x y
x x x ,
y y y µ x y
x x y .
y x y Both of these give associative operations, as we check by verifying
each of
(xx)y = xy = x(xy), (xy)x = x(yx), (yx)x = y(xx) = yx,
(xy)y = x(yy) = xy, (yx)y = y(xy), (yy)x = yx = y(yx). Next let us look for commutative associate operations. Here we know
that xy = yx, which implies (xx)x = x(xx) and (yy)y = y(yy), and
reduces the other requirements of associativity to
(xx)y = x(xy) = x(yx) = (xy)x = y(xx),
(yy)x = y(yx) = y(xy) = (yx)y = x(yy).
If we try xy = yx = x these become
(xx)y = xx = y(xx), (yy)x = x = x(yy), and we see that if xx = x then yy can be x or y, while if xx = y then
yy = y. Thus xy = yx = x yields 3 possible associative operations,
µ x y
x x x ,
y x x µ x y
x x x ,
y x y µ x y
x y x .
y x y Finally, if we impose commutativity with xy = yx = y, the associativity conditions become
(xx)y = y = y(xx), (yy)x = yy = x(yy) and so if yy = y then xx can be x or y, while if yy = x then xx = x.
Thus xy = yx = y yields 3 possible associative operations,
µ x y
x x y ,
y y y µ x y
x y y ,
y y y µ x y
x x y .
y y x So there are eight associative binary operations on X.
(2) Exercise 2 on page 12 of the notes
Solution. As in the previous exercise, we can picture a binary operation as a table that we need to fill in. In this case there are 100
spots in the table, each of which has 10 possible values. This gives
a total of 10100 binary operations on a set with ten elements.
(3) Exercise 4 on page 12 of the notes
Solution.
(a) First assume that f is surjective and let us show that there exists
a g satisfying f ◦ g = id. We define g by choosing for each r ∈ R a
number s = g(r) such that f (s) = r (we know that there is such an
s because f is surjective).
Next assume that there exists a g satisfying f â—¦ g = id and let us
show that f is surjective. Given r ∈ R we need to show that there
is an s ∈ R such that f (s) = r, but we know that f (g(r)) = r, so we
see that s = g(r) works.
(b) First assume that f is injective and let us show that there exists
a g satisfying g â—¦ f = id. We define g differently on the image of f
and on numbers that are not in the image of f. If r = f (s) for some
s (so, by injectivity of f, for a single s), then we define g(r) = s.
If r is not equal to f (s) for any s, then we can define g(r) to be
anything we like, let us say g(r) = 0 if r is not in the image of f. By
construction, this function g satisfies g(f (s)) = s for all s ∈ R.
Next assume that there exists a g satisfying g â—¦ f = id and let us
show that f is injective. We need to show that if f (s) = f (s0 ) then
s = s0 . If f (s) = f (s0 ) then g(f (s)) = g(f (s0 )) but g â—¦ f = id, so this
equality is s = g(f (s)) = g(f (s0 )) = s0 , which is what we wanted to
show. (c) No. If f is not injective, say f (a) = f (b) = y, and g is a right
inverse with g(y) = a then we can define a right inverse h =
6 g by
(
g(x) if x 6= y
h(x) =
b
if x = y
(4) Exercise 5 on page 12 of the notes
Solution. Let (X, ∗) be a monoid and x, y invertible elements. First
let us check that x−1 is invertible. The fact that x−1 is the inverse
of x means that
x ∗ x−1 = e and x−1 ∗ x = e,
but these equations also tell us that x−1 is invertible and exhibit x
as its inverse, i.e., as (x−1 )−1 .
Next let us check that x ∗ y is invertible and its inverse is y −1 ∗ x−1 .
This follows from the equations
(x ∗ y) ∗ (y −1 ∗ x−1 ) = x ∗ (y ∗ y −1 ) ∗ x−1 = x ∗ e ∗ x−1 = x ∗ x−1 = e
(y −1 ∗ x−1 ) ∗ (x ∗ y) = y −1 ∗ (x−1 ∗ x) ∗ y = y −1 ∗ e ∗ y = y −1 ∗ y = e,
where we have used associativity.
In additive notation, x−1 = −x, so the first statement becomes
−(−x) = x and the second −(x + y) = −y − x.
(5) Exercise 6 on page 12 of the notes
Solution.
(a) We prove xn ∗ xm = xn+m by induction on m. For m = 1 this is
true by definition. Assume that this is true for all m = ` − 1 and let
us show that it is true for m = `. This follows from
xn ∗ x` = xn ∗ (x`−1 ∗ x) = (xn ∗ x`−1 ) ∗ x = xn+`−1 ∗ x = xn+`
where the last equality is by definition and the previous one by inductive hypothesis. This completes the induction.
Next we prove (xn )m = xnm by induction on m. For m = 1 this is
true by definition. Assume that this is true for all m = ` − 1 and let
us show that it is true for m = `. This follows from
(xn )` = (xn )`−1 ∗ xn = (xn(`−1) ) ∗ x` = xn(`−1)+` = xn`
where the first equality is by definition, the second is inductive hypothesis, and the third uses the property xn ∗ xm = xn+m which we
just proved. This completes the induction.
(b) First let us show that x ∗ y = y ∗ x implies xn ∗ y = y ∗ xn for all
n ∈ N by induction on n. For n = 1 this is true by assumption. If it
holds for n = ` − 1 then let us show that it is true for n = `. This follows from
x` ∗ y = (x ∗ x`−1 ) ∗ y = x ∗ (x`−1 ∗ y) = x ∗ (y ∗ x`−1 ) = (x ∗ y) ∗ x`−1
= (y ∗ x) ∗ x`−1 = y ∗ (x ∗ x`−1 ) = y ∗ x` ,
where we have used associativity and the inductive hypothesis, which
completes the induction.
Next we prove xn ∗ y n = (x ∗ y)n for x and y satisfying x ∗ y = y ∗ x
by induction on n. For n = 1 this is true by definition. Assume it
is true for n = ` − 1 and let us show that it is true for n = `. This
follows from
x` ∗ y ` = (x ∗ x`−1 ) ∗ (y ∗ y `−1 ) = x ∗ (x`−1 ∗ y) ∗ y `−1 = x ∗ (y ∗ x`−1 ) ∗ y `−1
= (x ∗ y) ∗ (x`−1 ∗ y `−1 ) = (x ∗ y) ∗ (x ∗ y)`−1 = (x ∗ y)`
completing the induction.
In additive notation, xn = nx, and these results are expressed
nx + mx = (n + m)x, n(x + y) = nx + ny. (6) Exercise 1 on page 17 of the notes
Solution. The tables for (G, +) and (H, ×) define binary operations
which from exercise 1 are associative. We note that 0 is a unit for
(G, +) and 1 is a unit for (H, ×). These are necessarily their own
inverse, so we only need to check that the other element has an inverse. The inverse of 1 in (G, +) is 1 since 1 + 1 = 0 and the inverse
of −1 in (H, ×) is −1 since −1 × −1 = 1.
(7) Exercise 3 on page 17 of the notes
Solution. Let us check that g ∗ x = h has the unique solution
x = g −1 ∗ h. Indeed, multiplying both sides on the left by g −1 yields
g −1 ∗ h = g −1 ∗ (g ∗ x) = (g −1 ∗ g) ∗ x = e ∗ x = x.
On the other hand, if x0 is another solution then we have g ∗x = g ∗x0
but then the cancellation law tells us that x = x0 . The proof that
x ∗ g = h has the unique solution x = h ∗ g −1 is similar.
(8) Exercise 7 on page 17 of the notes
Solution. To show that C(G) is an Abelian subgroup, let us start
by showing that (C(G), ∗) is an algebraic structure. First C(G) 6= ∅
because the identity of G commutes with all elements of G and hence
is in C(G). Next, assume that x, y ∈ C(G), we need to show that
x ∗ y ∈ C(G). That is we need to check that x ∗ y commutes with all elements of G, so let z ∈ G and note that
(x ∗ y) ∗ z = x ∗ (y ∗ z) = x ∗ (z ∗ y) = (x ∗ z) ∗ y
= (z ∗ x) ∗ y = z ∗ (x ∗ y).
This shows that x ∗ y ∈ C(G) as required. Next, we know that
(C(G), ∗) is associative since (G, ∗) is associative and we know that
e ∈ C(G) is a unit for (C(G), ∗) since it is a unit for (G, ∗). This
shows that (C(G), ∗) is a monoid so next we show whenever x ∈
C(G) we also have x−1 ∈ C(G). Thus we need to show that x−1
commutes with any y ∈ G, but we have
x−1 ∗ y = (y −1 ∗ x)−1 = (x ∗ y −1 )−1 = y ∗ x−1 ,
and so x−1 ∈ C(G). Thus shows that (C(G), ∗) is a group. Finally,
if x, y ∈ C(G) then x ∗ y = y ∗ x since this is true for any y ∈ G,
hence (C(G), ∗) is an Abelian group. (Note that we could also have
used Proposition 1.2.5 to (slightly) shorten the argument.)
Next consider G = GLn (R). For every s 6= t consider the matrix Est
whose entries are all zero except for the (s, t) entry which is equal
to one. For example, when n = 3, we have 0 1 0
0 0 1
0 0 0
E12 = 0 0 0 , E13 = 0 0 0 , E21 = 1 0 0 , etc.
0 0 0
0 0 0
0 0 0
Notice that det(Id +Est ) = 1, so Id +Est in always in GLn (R). If A
is in the center of GLn (R) then we must have
A(Id +Est ) = (Id +Est )A ⇐⇒ AEst = Est A.
Let us compute the (i, j) entry of each side of this equality,
(
X
0
if j 6= t
(AEst )ij =
aik (Est )kj =
,
ais if j = t
k
(
X
0
if i 6= s
(Est A)ij =
(Est )ik akj =
atj if i = s
Since these have to be equal, we conclude that
ais = 0 if i 6= s, atj = 0 if j 6= t, and ass = att . Thus, since s and t were any two distinct indices, we can conclude
that if A is in the center of GLn (R) it must be a multiple of the
identity. On the other hand, any multiple of the identity is in the
center of GLn (R), so we have shown that
C(GLn (R)) = {λ Id : λ ∈ R}. (9) Exercise 8 on page 17 of the notes
Solution. By Proposition 1.2.5, we need to show that if x, y ∈
H1 ∩ H2 then x ∗ y −1 ∈ H1 ∩ H2 . However we know that x ∗ y −1 ∈ H1
since H1 is a subgroup and x, y ∈ H1 , and similarly that x∗y −1 ∈ H2 .
It follows that x ∗ y −1 ∈ H1 ∩ H2 , and hence that this is a subgroup.
(10) Exercise 9 on page 17 of the notes
Solution. First, if G is Abelian then we know from exercise 5 that
(g ∗ g 0 )2 = g 2 ∗ (g 0 )2 for all g, g 0 ∈ G. So we only need to show that
if this equality holds then G is Abelian. Given g, g 0 ∈ G, we have
g ∗ g 0 ∗ g ∗ g 0 = (g ∗ g 0 )2 = g 2 ∗ (g 0 )2 = g ∗ g ∗ g 0 ∗ g 0
using the cancellation property we can cancel the g on the left to get
g0 ∗ g ∗ g0 = g ∗ g0 ∗ g0.
Using the cancellation property again we can cancel the g 0 on the
right to get
g0 ∗ g = g ∗ g0
which shows that G is Abelian.
(11) Exercise 11 on page 18 of the notes
Solution. We know that (G, ∗) is an associative algebraic structure
(or semigroup).
Let us check that G has a unit. Pick g ∈ G and let a be a solution
of the equation g ∗x = g and b be a solution of the equation x∗g = g.
We will first show that a = b. Let h be a solution to g ∗ x = a and
let k be a solution to x ∗ g = b. Then we have
a = g ∗ h = (b ∗ g) ∗ h = b ∗ (g ∗ h) = b ∗ a
= (k ∗ g) ∗ a = k ∗ (g ∗ a) = k ∗ g = b.
We have shown that a = b. Note that this shows that there is only
one solution to g ∗ x = g and only one solution to x ∗ g = g and that
these solutions are the same. Let us denote this solution by eg .
If g 0 ∈ G then by the same argument there is a unique solution to
g 0 ∗ x = g 0 and to x ∗ g 0 = g 0 and we denote this solution by eg0 . We
will next show that eg = eg0 . Let i be a solution to g 0 ∗ x = eg and
let j be a solution to x ∗ g = eg0 . Then we have
eg = g 0 ∗ i = (eg0 ∗ g 0 ) ∗ i = eg0 ∗ (g 0 ∗ i) = eg0 eg
= (j ∗ g) ∗ eg = j ∗ (geg ) = j ∗ g = eg0 .
This shows that there is a single element solving g ∗ x = g and
x ∗ g = g for all g ∈ G. That is, we have shown there is a unit. We
will denote the unit by e.
Finally we need to show that there are inverses. We know that
every g ∈ G has a left inverse and a right inverse, so by a proposition proven in class, it is invertible.
(12) Exercise 12 on page 18 of the notes
Solution. We know that (G × H, ◦) is an algebraic structure. Let
us check that it is associative,
((g, h)◦(g 0 , h0 ))◦(g 00 , h00 ) = (g ∗g 0 , h·h0 )◦(g 00 , h00 ) = ((g ∗g 0 )∗g 00 , (h·h0 )·h00 )
= (g∗(g 0 ∗g 00 ), h·(h0 ·h00 )) = (g, h)◦(g 0 ∗g 00 , h0 ·h00 ) = (g, h)◦((g 0 , h0 )◦(g 00 , h00 )),
that is has a unit,
(g, h) ◦ (e, e) = (g ∗ e, h · e) = (g, h) and (e, e) ◦ (g, h) = (e ∗ g, e · h) = (g, h),
and that (g −1 , h−1 ) is the inverse of (g, h),
(g, h) ◦ (g −1 , h−1 ) = (g ∗ g −1 , h · h−1 ) = (e, e)
and (g −1 , h−1 ) ◦ (g, h) = (g −1 ∗ g, h−1 · h) = (e, e).
This shows that (G × H, ◦) is a group.
Next consider the case where G = H = R with addition. Elements
of (R × R, ◦) are pairs of real numbers with the operation
(a, b) â—¦ (c, d) = (a + b, c + d),
thus we have pairs of real numbers with addition in each coordinate,
which is precisely (R2 , +).

 


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