Download Solutions to coursework 10 File

MTH4104
Introduction to Algebra
Solutions 10
Semester B, 2017
Question 1 Let S be the set of all complex numbers of modulus 1. Prove that S is a
subgroup of the multiplicative group C× .
Solution Since S is a subgroup of C× , and we are implicitly supposing that the group
operation of S is the same as the group operation of C (namely multiplication of complex numbers), we can prove this using the subgroup test. Using the test, what we have
to show is that if z and w are elements of S, then so is zw−1 .
In week 3 of this module we learned that modulus of complex numbers is a multiplicative function, that is, that |zw| = |z| · |w| for any complex numbers z and w. It
follows, from
1 = |1| = |zz−1 | = |z| · |z−1 |,
that |z−1 | = 1/|z|. Therefore, if z and w are elements of S so that |z| = |w| = 1, we have
|zw−1 | = |z| · |w−1 | =
|z|
1
= = 1,
|w| 1
implying zw−1 ∈ S. By the subgroup test, we may conclude that (S, ·) is a subgroup
of C× .
Question 2
(a) Let g and h be elements of a group G. Prove that if gh = hg, then g−1 h = hg−1 .
(b) Let G be a group, and h an element of G. Let C be the set {g ∈ G : gh = hg}.
Describe C in words.
(c) Prove that C is a subgroup of G.
Solution (a) Let gh and hg be such elements. Multiply the equation gh = hg by g−1
on both sides:
hg−1 = g−1 ghg−1 = g−1 hgg−1 = g−1 h,
which is what was to be proved.
(b) A practicing algebraist would say it this way: C is the set of all elements of G that
commute with h. (Well, actually they’d use the special name for subgroups of this form
and call it the centraliser of h; but I certainly wasn’t expecting you to discover this
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word!) If two elements g and h satisfy the equation gh = hg, we say that “g and h
commute”, or “g commutes with h”.
(c) We use the subgroup test: what we must show is that if f and g are elements of C,
then so is f g−1 . We have
f g−1 h
= f hg−1
= hfg
−1
because g ∈ C, using part (a)
because f ∈ C
which proves that f g−1 ∈ C, completing the proof.
Question 3 Let m > 1 be a natural number. Let R be the set of all functions from Zm
to Zm , with addition defined by addition of values,
( f + g) : x 7→ f (x) + g(x)
for all x ∈ Zm ,
and multiplication defined as composition,
( f g) : x 7→ f (g(x))
for all x ∈ Zm .
Is R a ring with these operations?
Solution The structure R is not a ring, because it fails to satisfy one half of the distributive law. To wit, it is not always true that
f g + f h = f (g + h)
for f , g, h ∈ R.
Here is a counterexample that works for any m > 1. Let g and h both be the identity
function, while f is the function so that f (0) = 1 while f (x) = 0 for any nonzero x.
Then the values of f g + f h and f (g + h) are as in the following table:
x
( f g + f h)(x) = f (g(x)) + f (h(x)) = 2 f (x)
( f (g + h))(x) = f ((g + h)(x)) = f (2x)
0
2
1
nonzero
0
0
and we see that they are unequal functions, because [2]m 6= [1]m if m > 1.
As it turns out, R satisfies all the other axioms of rings with identity, including the
other half of the distributive law, f h + gh = ( f + g)h.
Question 4
order.
Write the following element of S11 in cycle notation, and calculate its
1 2 3 4 5 6 7 8 9 10 11
g=
.
5 8 2 9 6 10 7 11 1 4 3
2
Solution Follow the algorithm given in lectures: Open a bracket and put the first
number 1. Then find images until you come back to 1:
1 7→ 5 7→ 6 7→ 10 7→ 4 7→ 9 7→ 1,
so the cycle is completed, (1 5 6 10 4 9). We are not finished: next we open another
bracket with 2 and find images to complete the cycle (2 8 11 3). And there is one
element 7 left, which is a fixed point, i.e. a cycle of its own, (7). So
g = (1 5 6 10 4 9)(2 8 11 3).
It is also correct to insert the one-element cycle (7).
The order of a permutation is the least common multiple of its cycle lengths. In this
case, the lcm of 6, 4, 1 is 12.
Question 5 Let g be the permutation in the preceding question, and let
h = (1 10 3 9 7 6 11)(2 8)(4 5).
(a) Calculate h−1 , gh, and h−1 gh.
(b) What is the order of h−1 gh? Explain how and why the order of g is related to the
order of h−1 gh.
Solution
(a) For h−1 , just read the cycles backwards:
h−1 = (11 6 7 9 3 10 1)(8 2)(5 4).
For gh, the procedure to compute it directly in the cycle notation is as follows. Open
a bracket and put in the first number 1. Then
1 7→ 10 (by h) 7→ 4 (by g),
so write 4 next in the bracket. Then
4 7→ 5 (by h) 7→ 6 (by g),
so write 6 next in the bracket. Then
6 7→ 11 (by h) 7→ 3 (by g),
so write 5 next in the bracket. Then
3 7→ 9 (by h) 7→ 1 (by g),
which completes the cycle
(1 4 6 3).
We are not yet done with the product. Open another bracket and put in 2, the first
number not yet accounted for. Then
2 7→ 8 (by h) 7→ 11 (by g),
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so write 10 next in the bracket. Then
11 7→ 1 (by h) 7→ 5 (by g),
etcetera. This gives us the cycle
(2 11 5 9 7 10).
Finally, 8 is the smallest number unaccounted for, and tracing its images shows that it
is a fixed point. We now have all the elements accounted for and conclude
gh = (1 4 6 3)(2 11 5 9 7 10).
Adding the (8) is optional.
Of course, it would be also perfectly acceptable to convert g and h into the two-line
notation and compose them as usual.
For h−1 gh, multiply the two permutations just found by the same procedure:
h−1 gh = (1 5 3 11 4 7)(2 6 10 8)
with optional (9).
(b) The cycle lengths of h−1 gh are the same as those of g, namely 6, 4, 1, and thus they
are of the same order, 12.
This equality of orders is a general property of the elements g and h−1 gh in any
group. We call two such elements conjugate; you may have seen this word before in
the context of matrix multiplication.
Informally, the reason why g and h−1 gh have the same order here is because “h−1 gh
is just g with the elements of {1, . . . , n} given different names”: the “new name” of any
number a ∈ {1, . . . , n} is h−1 (a). Indeed,
h−1 gh(h−1 (a)) = h−1 ghh−1 (a) = h−1 g(a) = h−1 (g(a))
so all h−1 gh is doing is computing g(a) and accounting for this “renaming”.
If you wanted to prove this formally for all groups, the key property to establish
would be that for any natural number n, we have
(h−1 gh)n = h−1 gn h.
[Expand it out yourself and see! In fact this is actually true for any integer n, negatives
included.] So (h−1 gh)n becomes 1 for the first time at the same exponent n for which
h−1 gn h does. But the equation h−1 gn h = 1 is equivalent to gn = h1h−1 = 1, so this first
n is also the order of g.
Question 6 Let G be the symmetric group S3 .
(a) What is the order of G?
(b) What are the orders of the elements of G?
(c) Show that G contains just one subgroup of order 3 and three subgroups of order 2.
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Solution
(a) The order of S3 is 3! = 6, the number of permutations of {1, 2, 3}.
(b) The identity has order 1. The three transpositions (1, 2), (1, 3) and (2, 3) each have
order 2. The other two elements (1, 2, 3) and (1, 3, 2) each have order 3.
(c) By Lagrange’s Theorem, elements of a subgroup of order 2 must have order 1 or 2,
and the identity must obviously be included. So there are three possibilities: {e, (1, 2)},
{e, (1, 3)}, and {e, (2, 3)}. It is easy to see that each is a subgroup. Most easily: for any
element g of a group G, we know that hgi, the subgroup generated by g (which consists
of all the powers gn for n ∈ Z) is a subgroup, and its order is equal to the order of g. So,
for example, h(1, 2)i = {e, (1, 2)} is a subgroup of order 2.
Similarly a subgroup of order 3 must have elements of orders 1 or 3, and there is
only one possibility: {e, (1, 2, 3), (1, 3, 2)} (which again is a subgroup, as it is h(1, 2, 3)i).
Question 7 Does S6 contain
(a) an element of order 7?
(b) a subgroup of order 7?
(c) an element of order 8?
(d) a subgroup of order 8?
Justify your answers.
Solution (b) No , by Lagrange’s theorem: S6 cannot have a subgroup of order 7
because 7 does not divide |S6 | = 6! = 720.
(a) No , because if there were an element g ∈ S6 of order 7, then the cyclic subgroup
hgi it generates would be a subgroup of S6 of order 7, and this is ruled out by part (b).
(c) The order of an element g ∈ S6 is the least common multiple of the lengths of all its
cycles, when g is written in cycle notation. So, if g is to have order 8, its cycle lengths
must have lcm 8. However, the sum of these cycle lengths is at most 6, since g only has
the six elements {1, 2, 3, 4, 5, 6} to act on. And 8 is not the lcm of any naturals less than
or equal to 6: because 8 = 23 is a power of a prime, it will only show up as the lcm of
a collection of integers if one of them was already divisible by 23 . So no , there is no
such g.
(d) Since 8 is a divisor of 6!, Lagrange’s theorem permits the existence of a subgroup
of order 8 of S6 . But it does not guarantee that one exists—the converse of Lagrange’s
theorem is false—so this approach is inconclusive.
Part (c) tells us that a cyclic subgroup of S6 cannot have order 8. But perhaps there
is a non-cyclic subgroup that does.
In fact, yes , there is such a subgroup. One example is
{e, (1 2), (3 4), (1 2)(3 4), (5 6), (1 2)(5 6), (3 4)(5 6), (1 2)(3 4)(5 6)}.
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How did I find this? More or less, by a variation of the argument we used to compute
orders of elements in cycle notation. If g and h are disjoint cycles, and g has order m
while h has order n, then gh has order lcm(m, n): powers of gh can be written
gh, g2 h2 , g3 h3 , . . .
because g and h commute, and the first time I get gi hi = e in this list is when both
gi = e and hi = e. What if I wasn’t increasing the exponents on g and h in lockstep,
but allowed them to vary separately? Then there would be m different values taken by
gi , and n different values taken by h j , and any product of one of the former m elements
with one of the latter n would give me a different result, for mn results overall. It’s
also not hard to check this is a subgroup [try it yourself!] So by packing three cycles
of length 2 into our 6-element set, and allowing arbitrary exponents on each of them, I
produce 2 · 2 · 2 = 8 different elements overall.
Question 8
(a) Prove that the set of permutations f of {1, . . . , n} such that f (n) = n is a subgroup
of Sn .
(b) Describe the cosets of this subgroup, in as simple a fashion as you can.
Solution (a) Give the name H to this subset,
H = { f ∈ Sn : f (n) = n}.
We will use the Subgroup Test to show that H is a subgroup. This means that we must
show that if g and h are in H, then also gh−1 = H.
So let g and h be permutations of {1, . . . , n} such that g(n) = h(n) = n. Since
h(n) = n, the element that maps to n under h is n, that is, h−1 (n) = n. It follows that
(gh−1 )(n) = g(h−1 (n)) = g(n) = n
so that gh−1 ∈ H. QED.
(b) Any coset of H can be written in the form Hg = {hg : h ∈ H} for some element g
of Sn .
In order to make this description simpler, consider what we know about all these
permutations hg. Since h(n) = n for all h ∈ H, if a is the element of {1, . . . , n} such that
g(a) = n, then also
(hg)(a) = h(g(a)) = h(n) = n
for every h ∈ H. That is, a is the element mapped to n by every permutation in Hg.
Actually Hg is exactly the set of permutations f such that f (a) = n: one way to see this
is that it consists only of such permutations but its size must be |Hg| = |H| = (n − 1)!,
and therefore it must contain all (n − 1)! of the permutations that map a to n.
In summary, the cosets of H are the n sets
{ f ∈ Sn : f (a) = n}
for each a = 1, 2, . . . , n.
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Question 9 Let f : X → Y be a function. In lecture I asserted that f has an inverse
function if and only if it is a bijection. This is a corollary of the following:
(a) Prove that there exists a function g such that f ◦ g is the identity function on Y if
and only if f is a surjection.
(b) Prove that there exists a function h such that h ◦ f is the identity function on X if
and only if f is an injection.
(c) Prove that, if f is a bijection, then the g and h from parts (a) and (b) can be chosen
to be equal to each other.
[Feel free to assume that the sets X and Y are finite.]
Solution (a) Suppose that ( f ◦ g)(y) = y for all y ∈ Y . Then for all y ∈ Y , there is an
x ∈ X such that f (x) = y: indeed, x = g(y) works, because f (g(y)) = y by assumption.
So f is surjective.
Conversely, suppose f is a surjection. Then we can define g as follows: for each
y ∈ Y , pick any one of the elements x ∈ X such that f (x) = y, and let g(y) = x. Clearly
this definition ensures that f (g(y)) = y for each y ∈ Y .
(b) Suppose that (h ◦ f )(x) = x for all x ∈ X. Let x1 and x2 be different elements of x;
we wish to show that f (x1 ) and f (x2 ) are also different. But if they weren’t, we would
have
x1 = h( f (x1 )) = h( f (x2 )) = x2 ,
contradicting that x1 and x2 were chosen to be different. So f is injective.
Conversely, suppose f is an injection. Then we can define h as follows. For y ∈ Y , if
y = f (x) for some x, then this x is unique; let h(y) = x. Otherwise, if y is not in the range
of f , let h(y) take any arbitrary value. Clearly this definition ensures that h( f (x)) = x
for each x ∈ X.
(c) This is an immediate consequence of our constructions above. Both g and h are the
function that is completely defined by saying that it maps f (x) back to x for each x. So
g and h are equal. (Despite the wording of the question, there is not actually any choice
in what g or h is in this situation.)
Commentary. The finiteness assumption is not necessary for this question. I allowed
it for the following reason. The way I constructed g in part (a) relies on the axiom of
choice, which says that, given an (infinite) family of sets, you can choose an element
arbitrarily from each one. As innocuous as this axiom may seem, it is actually independent of the remaining axioms of standard (Zermelo-Fraenkel) set theory, i.e. you
cannot prove it from the remaining axioms! Zermelo introduced the axiom of choice
in 1904, and for the better part of the twentieth century it was controversial, with many
set theorists not accepting it. Nowadays it is generally accepted. All the same, given
my emphasis on the axiomatic method, it didn’t seem right to ask you to construct a
function in a way that isn’t actually allowed under a historically popular set of axioms
for set theory.
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