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D-MATH
Prof. Brent Doran
Algebra I
HS 2013
Solution 3
Modular arithmetic, quotients, product groups
1. Show that the functions f = 1/x, g = (x − 1)/x generate a group of functions,
the law of composition being composition of functions, that is isomorphic to the
symmetric group S3 .
Solution The function f has order 2 in the law of composition of functions, in
fact
f (f (x)) = 1/f (x) = 1/(1/x) = x,
similarly the function g has order 3:
1
= ((x−1)/x)−1
= − x−1
g 2 (x) = g(x)−1
g(x)
(x−1)/x
1
1
1
g 3 (x) = g 2 (g(x)) = − g(x)−1
= − (x−1)/x−1
= − −1/x
=x
Moreover the equality f ◦ g 2 = g ◦ f holds:
1 − x = f ◦ g2 = g ◦ f =
1/x − 1
= x(1 − x)/x = 1 − x.
1/x
This implies that the group generated by f and g has the same multiplication
table as S3 , hence the homomorphism defined by sending f to (12) and g to (123)
gives an isomorphism of the two groups.
2. In the additive group Rm of vectors, let W be the set of solutions of a system
of homogeneous linear equations AX = 0. Show that the set of solutions of an
inhomogeneous system AX = B is either empty, or is an additive coset of W .
Solution A vector space V is an abelian group that is endowed with the additional
structure of scalar multiplication. Any linear map between vector spaces is a group
homomorphism, that is also compatible with the additional structure. Let us now
fix the matrix A representing an endomorphism of the vector space V . And let
W < V denote the image of A, K its kernel.
The First Isomorphism Theorem says that a surjective map A : V → W whose
kernel is K, induces a group isomorphism V /K → W . Since this is the case, the
space W corresponds bijectively to the left cosets of K. Indeed since the group V
is abelian, left cosets and right cosets are equal and any subgroup of V is normal.
The inhomogeneous equation AX = B admits solution if and only if the vector B
belongs to the image W , and in that case the set of solutions consist of the vectors
that can be written as X0 + v where v is in the kernel K of A. This shows that
the solutions of AX = B are left cosets of K and any left coset of K correspond
to the solution of AX = B for an element B in W .
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3. Prove that every subgroup of index 2 is a normal subgroup. Find an example of a
subgroup of index 3 that is not normal.
Solution Let N be a subgroup of index two in G. Since the left cosets of N form
a partition of G consisting of exactly two elements, the coset of N distinct from
N consists of all the elements of G that are not in N . The same is true for the
right cosets, that, again, are N and N c .
We want to show that for any element g ∈ G, gN g −1 = N . We distinguish two
cases. Let us assume first that g belongs to N . In that case for any element n ∈ N
the element gng −1 is still an element of N since N is a group, hence gN g −1 = N .
Let us now assume that g doesn’t belong to N , then the left coset gN is not N
since it contains the element g that, by assumption, doesn’t belong to N and hence
consists of all the element of G that are not in N . The same is true for the right
coset, in particular we get that gN = N g. This implies that for any n in N there
is n0 ∈ N such that gn = n0 g multiplying on the right g −1 we get gng −1 = n0 and
this implies that gN g −1 = N .
In order to give an example of a subgroup of index 3 that is not normal, let us
consider the symmetric group S3 and the subgroup H generated by the permutation h = (12). The group S3 has order 6, and the group H has order 2, since
the transposition (12) has order 2. In order to show that H is not normal let us
consider the permutation g = (13):
ghg −1 = (13)(12)(13) = (132).
Since ghg −1 do not belong to H, the subgroup H is not normal.
It is interesting to compute the left and the right cosets of the group H and see
that they give two different partitions of G:
Left cosets
eH = {e, (12)}
(13)H = {(13), (123)}
(23)H = {(23), (132)}
Right cosets
He
= {e, (12)}
H(13) = {(13), (132)}
H(23) = {(23), (123)}.
4. Let H < G be a finite index subgroup, then there exist a subgroup N of H that
is normal in G and has finite index.
Solution
Let us denote by n the index of H in G and let eH = g1 H, g2 H, . . . gn H be the
left cosets of H in G. Let us now consider the intersection
N=
n
\
gi Hgi−1 .
i=1
N is a subgroup of H, since the intersection of groups is a group and since g1 H = H
hence g1 belongs to H and g1 Hg1−1 = H.
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Let us now check that N is normal in G. Let g be an element of G. The left
multiplication by g induces a permutation of the left cosets of H, in particular
there is a permutation σ ∈ Sn and elements hi in H such that ggi = gσ(i) hi . This
−1
gives that g(gi Hgi−1 )g −1 = gσ(i) Hgσ(i)
, hence gN g −1 = N .
To show that N is finite index we will prove the following Lemma:
Lemma: If A and B are subgroups of index a and b in G, then A ∩ B is a subgroup
of index at most ab.
Proof: Indeed the cosets of A ∩ B give a partition of G, but since A ∩ B is a
subgroup of A the left cosets of A ∩ B that are contained in A give a partition of
A. We claim that this partition has at most b elements: indeed the cosets of B in
G give a partition of G consisting of b elements and (gB) ∩ A is either empty or
a coset of A ∩ B in A. Since the same is true for any coset hA of A in G, we have
that for any coset of A there are at most b cosets of A ∩ B covering it, and since
there are a cosets of A in G there are at most ab cosets of A ∩ B in G. End of the
proof.
Notice that the index of A ∩ B can be smaller then ab, this happens for example
when A = B.
T
By induction this implies that the intersection H = ni=1 Hi of the finite index
subgroups Hi of index hi has index at most h1 · . . . · hn , in particular has finite
index.
This applies in particular to N that is hence a normal subgroup of G of finite
index that is contained in H.
5. Determine the integers n for which the pair of congruences 2x−y ≡ 1 and 4x+3y ≡
2 modulo n has a solution.
Solution If the system of equations
2x − y ≡ 1
4x + 3y ≡ 2
are satisfied modulo n, we get, subtracting two times the first equation to the
second one that 5y ≡ 0 modulo n.
This implies that n must be odd. Indeed let us assume by contradiction that n
is even and there is a solution for both equations, we get from the first equation
that y must be odd, but since 5y must be zero, y must be even that gives a
contradiction.
We will now show that if n is odd there is always at least a solution. Indeed, since
n is odd, 2 is invertible modulo n, this implies that there exists a number x such
that 2x = 1 + n. The pair (x, n) solves both equations modulo n.
6. Prove the Chinese Remainder Theorem: let a, b, u, v be integers, and assume that
the greatest common divisor of a and b is 1. Then there is an integer x such that
3
x ≡ u modulo a and x ≡ v modulo b. Find a counterexample if a and b are not
coprime.
Solution Let us consider first the case u = 1, v = 0. In this case we look for a
number x that satisfies x = 1 − am and x = bn. Finding x amounts to find integers
m and n such that 1 = am + bn. This problem is known as Bezout’s Lemma.
Bezout’s Lemma: if a and b are coprime there exist integers m and n such that
an + bm = 1.
Proof of Bezout’s Lemma: let us consider the smallest positive integer d that can be
written as an+bm for a pair of integers n, m. We want to show that d = 1. Assume
by contradiction that d > 1. Let n and m be numbers such that d = na+mb. Since
a and b are coprime, and d is bigger than 1, there is at least one between a and b
that is not exactly divisible by d, let us assume that a is such number and let us
consider the reminder c of the division of a by d. The number c is strictly smaller
than d, but positive moreover c = a − kd = a − k(na + mb) = (1 − kn)a + kmb.
And this gives a contradiction. End of the proof.
Let us now consider the element x = ubm + van, where m and n are integers given
by Bezout’s Lemma. We claim that x is a solution for the problem.
Indeed x = u(1 − an) + v(an) = u + a(vn − un) ≡ u modulo a. Similarly x =
v(1 − bm) + u(bm) = v + bm(u − v) ≡ v modulo b.
7. In each of the following cases determine whether or not G is isomorphic to the
product group H × K
(a) G = R× , H = {±1}, K = {positive real numbers}.
Solution Yes: indeed R× is abelian, both K and H are normal subgroups,
obviously they intersect only in 1, that is the identity of R× , and any real
number can be written as hk with h ∈ H, k ∈ K.
(b) G = {invertible upper triangular 2×2 matrices}, H = {invertible diagonal matrices}, K =
{upper triangular matrices with diagonal entries 1}.
Solution No: we will first show that the natural map
α: H ×K → G
(h, k) 7→ hk
doesn’t give a group isomorphism. Indeed if that was true, for any element
h in H and k in K, since by the group law in the product group we have
(0, k)(h, 0) = (h, k), we would get that kh = α(0, k)α(h, 0) = α(h, k) = hk.
But this is not true, indeed, for example
2 0 1 2
2 4
2 2
1 2 2 0
=
6=
=
.
0 1
0 1 0 1
0 1 0 1
0 1
Anyway it would still be possible that H × K was isomorphic to G under a
different isomorphism. For example, if G = Z2 xZ2 and H = K = Z2 x0 then
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HxK and G are isomorphic but not under the natural isomorphism given
by inclusion of H and K in G, since in that case the inclusions induce an
homomorphism that is not injective nor surjective.
We will now show that this is not the case in the problem 7b): the map
β : R× × Rt imes → H a 0
(a, b)
7→
0 b
a 0
gives a group isomorphism: the subgroups H0 = {
|a ∈ R× } and
0
1
1 0
H1 {
|a ∈ R× } are normal subgroups of H whose product is the whole
0 a
H and intersect in the trivial group.
The group K, instead is isomorphic to R+ : it is easy to check that the map
β : R+ → K 1 a
a 7→
0 1
gives a group isomorphism.
Let us now assume by contradiction that there exists an abstract isomorphism
H ×K ∼
= G. Then we would also get an isomorphism G ∼
= H0 × H1 × K ∼
=
×
×
+
×
+
R × R × R . Since both R and BR are abelian and the product of
abelian groups is abelian we would get that the group G would be abelian.
But this is not true and gives a contradiction.
(c) G = C× , H = {unit circle}, K = {positive real numbers}.
Solution Yes: since C× is abelian, both K and H are normal subgroups.
Moreover the two subgroups intersect only in the identity of C× . The fact
that the product KH is the whole C follows from the fact that every complex
number has an expression in polar coordinates, that is as a product of a
positive real number and a number in the unit circle.
8. Let H = {±1, ±i} be the subgroup of C× of fourth roots of unity. Describe the
cosets of H in G explicitly. Is G/H isomorphic to G?
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