Math 113 Homework 2
Solutions
due July 5, 2011
1. Prove that the additive group R+ of real numbers is isomorphic to the
multiplicative group P of positive reals.
Solution: Consider the function φ : R+ → P given by φ(x) = ex . φ is
a bijection, as it has an inverse φ−1 : P → R+ given by φ−1 (x) = ln(x).
Further, for all a, b ∈ R we have φ(a + b) = ea+b = ea eb = φ(a)φ(b), so φ
is a homomorphism. Hence φ is an isomorphism and R+ ∼
= P.
2. Prove that the matrices
1
1
1
A :=
and
B :=
1
1
1
are conjugate elements in the group GL2 (R), but that they are not conjugate when regarded as elements of SL2 (R).
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Solution: Consider the permutation matrix P =
. Observe that
1
P AP −1 = B. Further, det(P ) = −1, so P ∈ GL2 (R); this shows that A
and B are conjugate as elements of GL2 (R).
Now suppose they were conjugate
by some element Q ∈ SL2 (R), say
a
b
QAQ−1 = B, where Q =
. Then
c d
a a+b
a
b
QA =
= BQ =
c c+d
a+c b+d
Therefore a = 0 and b = c, so det(Q) = −b2 < 0, contradicting Q ∈
SL2 (R). So A and B are not conjugate as elements of SL2 (R).
3. Prove that the set Aut(G) of automorphisms of a group G forms a group,
the law of composition being composition of functions.
Solution: The automorphisms form a subset of the permutations of the
underlying set of G. Function composition is always associative, bijections
are invertible (with inverse map also a bijection), and a composition of
bijections is a bijection, so the permutations of the underlying set of G
form a group. We must show that Aut(G) is a subgroup.
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The identity map is clearly an automorphism.
If φ ∈ Aut(G), then for all a, b ∈ G we have
φ(φ−1 (ab)) = ab = φ(φ−1 (a))φ(φ−1 (b)) = φ(φ−1 (a)φ−1 (b))
as φ is a homomorphism; applying φ−1 to both sides we see that φ−1 is
likewise a homomorphism, hence an automorphism.
Finally if φ, ψ ∈ Aut(G), then for all a, b ∈ G, we have
φ(ψ(ab)) = φ (ψ(a)ψ(b)) = φ(ψ(a))φ(ψ(b))
so a composition of automorphisms is an automorphism, and we are done.
4. Determine the group of automorphisms of the following groups.
(a) Z+
Solution: Homomorphisms φ with cyclic domain are determined by
the image of a generator x, since then φ(xn ) = φ(x)n ; if the domain
is Z+ , 1 may go to any element of the codomain, so the set of homomorphisms φ : Z+ → Z+ (called the endomorphisms of Z+ ) is just
the collection of maps given by multiplication by an element of Z.
Of these, only 1 and −1 (i.e. the maps φ(x) = x, φ(x) = −x) are
bijections. Hence the group of automorphisms is {1, −1}× ∼
= Z/2Z.
More simply, a map of cyclic groups is a surjection iff the image of a
generator is a generator; Z has two generators {1, −1}, which may be
transposed or not.
(b) a cyclic group of order 10
Solution: Again, representing the group operation as addition modulo 10, endomorphisms of this group are given by multiplication by
some integer modulo 10. The only ones that are bijections are those
taking 1 to a generator, corresponding to multiplication by an integer
relatively prime to 10, namely 1, 3, 7, or 9. Thus |Aut(Z/10Z)| = 4.
Let φ ∈ Aut(Z/10Z) be the map given by ’multiplication by 3 modulo
10’. As 32 = 9 6= 1 modulo 10, φ has order larger than 2, hence equal
to 4, in Aut(Z/10Z), showing that Aut(Z/10Z) ∼
= Z/4Z is generated
by φ.
(c) S3
Solution: Given any group G, there is a homomorphism ψG : G →
Aut(G) taking an element a ∈ G to the automorphism of G given by
conjugation by a. Observe that an element a ∈ G is in the kernel of
this homomorphism iff conjugation by a is the identity map, that is,
aga−1 = g for every g ∈ G, that is, a ∈ Z(G). Since Z(S3 ) is trivial,
ψS3 : S3 → Aut(S3 ) is injective. This shows that |Aut(S3 )| ≥ 6.
On the other hand, S3 can be generated by the transpositions. Since
an automorphism preserves orders of elements, every automorphism
of S3 must permute the 3 transpositions of S3 . Since a homomorphism
is determined by its restriction to any set generating the domain, this
implies there are at most 3! = 6 automorphisms of S3 . Hence there
are exactly 6 and the above map ψS3 is an isomorphism, that is,
Aut(S3 ) ∼
= S3 .
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5. Prove that the kernel and image of a homomorphism are subgroups.
Solution: Let φ : G → H be a group homomorphism.
φ(1) = φ(1 · 1) = φ(1)φ(1)
so φ(1) = 1. This shows both that 1 ∈ ker(φ) and that 1 ∈ im(φ).
Further, for all a in G we have that 1 = φ(1) = φ(aa−1 ) = φ(a)φ(a−1 );
multiplying on the left by φ(a)−1 we obtain φ(a−1 ) = φ(a)−1 . This shows
that if a ∈ ker(φ) then a−1 ∈ ker(φ), and that if a ∈ im(φ) then a−1 ∈
im(φ).
Finally, if a, b ∈ ker(φ) then φ(ab) = φ(a)φ(b) = 1, so ker(φ) is closed and
hence a subgroup of G. Likewise, for any φ(a), φ(b) ∈ im(φ), φ(a)φ(b) =
φ(ab) shows that im(φ) is closed and hence a subgroup of H.
6. Describe all homomorphisms φ : Z+ → Z+ , and determine which are
injective, which are surjective, and which are isomorphisms.
Solution: As described in (4a), such a map φ is determined by the image
of 1, since it must be that φ(n) = nφ(1), and all of these are distinct. So
the maps are just multiplication by n for each n ∈ Z. These are injective
unless n = 0, but only surjective in the cases n = 1 or n = −1, which are
thus also the bijective cases.
7. (a) Let H be a subgroup of G, and let g ∈ G. The conjugate subgroup
gHg −1 is defined to be the set of all conjugates ghg −1 , where h ∈ H.
Prove that gHg −1 is a subgroup of G.
Solution: Let ghg −1 ∈ gHg −1 , where h ∈ H; then (ghg −1 )−1 =
gh−1 g −1 is a conjugate of h−1 ∈ H by g −1 , and so is also in gHg −1 .
Also if gh1 g −1 and gh2 g −1 are any two elements of gHg −1 , then
gh1 g −1 gh2 g −1 = g(h1 h2 )g −1 ∈ gHg −1 . Finally, 1 = g1g −1 ∈ gHg −1 ,
so gHg −1 is nonempty and thus a subgroup.
(b) Prove that a subgroup H of a group G is normal if and only if gHg −1 =
H for all g ∈ G.
Solution: If this equation holds, then for any h ∈ H, g ∈ G we
have ghg −1 ∈ gHg −1 = H, so H is normal in G. Conversely, if H is
normal, then for any g ∈ G, if x ∈ gHg −1 , there exists h ∈ H such
that x = ghg −1 = h, so in fact gHg −1 = H.
8. Prove that the center of a group is a normal subgroup.
Solution: If G is a group, then for all a ∈ Z(G), g ∈ G we have
gag −1 = agg −1 = a ∈ Z(G)
Therefore Z(G) is normal in G.
9. Prove that if a group contains exactly one element of order 2, then that
element is in the center of the group.
Solution: Let a be the unique element of order 2 in a group G. Then
as conjugation preserves order (by the last homework), for every g ∈ G
ag = gag −1 also has order 2, and hence ag = a. This means that ga = ag
for every g ∈ G, i.e. that a ∈ Z(G).
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10. Let φ : G → G0 be a surjective homomorphism, and let N be a normal
subgroup of G. Prove that φ(N ) is a normal subgroup of G0 .
Solution: Let n0 ∈ φ(N ) and g ∈ G0 . There exist n ∈ N and g ∈ G such
that φ(n) = n0 and φ(g) = g 0 . Then
g 0 n0 (g 0 )−1 = φ(g)φ(n)φ(n)−1 = φ(gng −1 ) ∈ φ(N )
since gng −1 ∈ N by normality of N . Thus φ(N ) is a normal subgroup of
G0 .
11. Let G be a group. Prove that the inversion function φ : G → G given by
φ(x) = x−1 is a homomorphism iff G is abelian.
Solution: Observe that for all a, b ∈ G, we have φ(ab) = φ(b)φ(a). Now,
if φ is a homomorphism, then for every a, b ∈ G we have
ab = φ(a−1 )φ(b−1 ) = φ(a−1 b−1 ) = φ((ba)−1 ) = ba
and therefore G is abelian. Conversely, if G is abelian, for every a, b ∈ G
we have
φ(ab) = (ab)−1 = b−1 a−1 = a−1 b−1 = φ(a)φ(b)
so φ is a homomorphism.
12. Prove that the nonempty fibres of a map form a partition of the domain.
Solution: Let f : A → B be a function. For each a ∈ A, a ∈ f −1 [f (a)],
so every element of the domain is in some nonempty fibre of f . Further,
if a ∈ f −1 (b), then b = f (a), so each element of the domain is in a unique
nonempty fibre of f . Hence the nonempty fibres of f form a partition of
A as desired.
13. (a) Prove that the relation x conjugate to y in a group G is an equivalence
relation on G.
Solution: Reflexivity: for every x ∈ G, x = x1 , so x ∼ x.
Symmetry: if x ∼ y there exists some g ∈ G such that xg = y, but
then x = yg−1 , so y ∼ x.
Transitivity: if x = yg and y = zh , then x = zhg , so x ∼ y and y ∼ z
implies x ∼ z.
(b) Describe the elements a whose conjugacy class (= equivalence class)
consists of the the element a alone.
Solution: An element a is of this type precisely when gag −1 = a for
all g ∈ G, that is, when a ∈ Z(G).
14. Prove directly that distinct cosets do not overlap.
Solution: Let H be a subgroup of a group G, let a, b ∈ G, and let
x ∈ aH ∩ bH. Then there are h1 , h2 ∈ H such that ah1 = x = bh2 .
−1
Multiplying on the left by h−1
1 , we obtain a = bh2 h1 . Then for any
y ∈ aH, there exists some h3 ∈ H such that y = ah3 = bh2 h−1
1 h3 ∈ bH, so
aH ⊂ bH. By symmetry, bH ⊂ aH, so aH = bH. Therefore, overlapping
cosets are identical, i.e. distinct cosets do not overlap.
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15. Let H, K be subgroups of a group G of orders 3, 5 respectively. Prove that
H ∩ K = {1}.
Solution: An intersection of subgroups is again a subgroup; this follows
because for all a, b ∈ H ∩ K, ab−1 ∈ H and ab−1 ∈ K, so ab−1 ∈ H ∩ K,
and further 1 ∈ H ∩ K. Thus H ∩ K is a subgroup both of H and of
K. By Lagrange’s theorem |H ∩ K| divides |H| = 3, and |H ∩ K| divides
|K| = 5, so |H ∩ K| = 1; therefore, H ∩ K = {1}.
Note that this argument immediately generalizes to show that any family
of subgroups such that the greatest common divisor of their orders is 1
has trivial intersection.
16. (a) Prove that every subgroup of index 2 is normal.
Solution: Let H be a subgroup of index 2 in a group G. Then the
left cosets of H in G must consist of H, and G\H. These must also
be the right cosets of H in G by the same argument. Therefore, the
left and right cosets of H coincide, i.e. gH = Hg for all g ∈ G. This
is equivalent to gHg −1 = H for all g ∈ G, which by (6b) is equivalent
to normality of H.
(b) Give an example of a subgroup of index 3 which is not normal.
Solution: Consider h(12)i ≤ S3 . It is an index 3 subgroup, but is
not normal as, for instance, (12)(23) = (13) ∈
/ h(12)i.
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