Mathematics 2F - Groups, Transformations, and Symmetries
Transformations, and Symmetries
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Mathematics 2F - Groups, Transformations, and Symmetries
Any electronic calculator may be used.
Candidates must attempt all questions.
1. Please complete the Moodle quiz on the Exam Moodle Page. Justification of
answers is not required for the Moodle quiz. 30
Each MCQ is worth 2 points.
2. Let M2×2(R) denote the set of 2×2 real matrices. Recall that M2×2(R) is a vector
space over R.
(i) Let S ⊂ M2×2(R) be the subset of matrices whose trace is zero. Prove that
S is a subspace. 4
The trace map Tr : M2×2(R)→ R is linear transformation. The set S is precisely
the kernel of Tr, so S is a subspace.
(ii) Compute the dimension of S. 3
Because the codomain of the trace map Tr : M2×2(R) → R is a 1-dimensional
vector space, the rank of Tr can only possibly be 1 or 0. Because Tr is not the
zero map, it cannot have rank 0. So it has rank 1. We have dimM2×2(R) = 4. By
the Rank-Nullity Theorem, we conclude that S = ker(Tr) has dimension 3.
(iii) Let U ⊂M2×2(R) be the subset of matrices whose determinant is zero. Prove
that U is NOT a subspace. 3
Consider the matrices
[
1 0
0 0
]
,
[
0 0
0 1
]
∈ U . Their sum
[
1 0
0 1
]
/∈ U . Therefore, U
is not a subspace.
3. Let G be a group and let ϕ : G→ GLn(Q) be a group homomorphism.
(i) Show that for n ∈ Z>0, the vector space Qn with the standard basis is a G-set
under the action
g.v = ϕ(g)v
[CONTINUED OVERLEAF
2for all g ∈ G and v ∈ Qn. 3
Denote by e ∈ G the identity in G. We have e.v = ϕ(e)v = In · v = v, where
In in the (n × n) identity matrix over Q. Let now g, h ∈ G. Then g.(h.v) =
ϕ(g)(ϕ(h)v) = (ϕ(g)ϕ(h))v = ϕ(gh)v = (gh).v.
(ii) Show that H = ϕ−1(SLn(Q)) is a normal subgroup of G. 3
First note that the preimage of any subgroup of the target of a group homomor-
phism is a subgroup of the domain. Hence H is a subgroup of G. To show that
it is normal, consider the group homomorphism det : GLn(Q) → Q∗. We have
H = ϕ−1(SLn(Q)) = ϕ−1(det−1(1)) = (det ◦ϕ)−1(1) = ker det ◦ϕ. Hence H can
be expressed as the kernel of a group homomorphism, and thus must be a normal
subgroup of G.
(iii) Let H be as in part (ii). Show that if ϕ is surjective, then we have an
isomorphism of groups G/H ∼= Q∗, where Q∗ = Q\{0} is viewed as a group under
multiplication. 4
By the previous part, we know that H = ker(det ◦ϕ). Hence, by the first iso-
morphism theorem, we have G/H ∼= det ◦ϕ(G). Now, since ϕ is surjective, we
have ϕ(G) ∼= GLn(Q), and hence G/H ∼= det(GLn(Q)). Now, the group homo-
morphism det : GLn(Q) → Q∗ is also surjective (for a ∈ Q∗ choose for exam-
ple the (n × n)-matrix A over Q with a as the top-left entry, 1s on the rest of
the diagonal, and zeroes everywhere else. Then det(A) = a). Hence we obtain
G/H ∼= det(GLn(Q)) = Q∗.
4. Let G be a group, and consider the set
Hom(Z, G) = {ϕ : Z→ G | ϕ is a group homomorphism}.
(i) Show that there is a bijection, i.e. an isomorphism of sets Hom(Z, G) ∼= G. 4
Consider the map ϕ : G→ Hom(Z, G), g 7→ ϕg, where ϕg : Z→ G, with ϕg(m) =
gm. This is well-defined: if g = h, then ϕg = ϕh, and ϕg ∈ Hom(Z, G): Indeed, ϕg
is a well-defined map from Z to G: Well-definedness is immediate from definition,
and we see that the image of ϕg lands in G, since for g ∈ G, we have gm ∈ G, as
G is closed under multiplication, has an identity and every element in G has an
inverse.
It remains to check that ϕ is a bijection. We first observe it is injective: Assume
that ϕg = ϕh. Then, in particular g = ϕg(1) = ϕh(1) = h, and hence ϕ is
injective. Finally, we show that ϕ is surjective: Let f ∈ Hom(Z, G) be a group
homomorphism from Z to G. Then for all m ∈ Z we have f(m) = f(1 + . . .+ 1︸ ︷︷ ︸
m times
) =
f(1) · · · f(1)︸ ︷︷ ︸
m times
= f(1)m. Therefore, we have that f = ϕf(1), and thus f lies in the
image of ϕ.
[CONTINUED OVERLEAF
3(ii) How many group homomorphisms are there from S3 to Z/17Z? Justify your
answer. 4
There is precisely one group homomorphism from S3 to Z/17Z: The trivial map
f : S3 → Z/17Z, σ 7→ 0 + 17Z, mapping every element in S3 to the identity.
Indeed, let ϕ : S3 → Z/17Z be a group homomorphism. Then its image ϕ(S3) is
a subgroup of Z/17Z. By Lagrange’s theorem, its order |ϕ(S3)| divides the order
of Z/17Z, where |Z/17Z| = 17. The only candidates for ϕ(S3) thus are the trivial
subgroup {0 + 17Z} and the group Z/17Z itself. If ϕ(S3) = Z/17Z, then ϕ is
surjective; a contradiction by the pigeonhole principle, since 6 = |S3| < |Z/17Z| =
17. Therefore, we must have ϕ(S3) = {0 + 17Z} and thus ϕ is the trivial map.
(iii) Define two different group homomorphisms from Z/2Z × Z/2Z to GL2(R).
Take for example f1 : Z/2Z × Z/2Z → GL2(R), (a, b) 7→ I2; the trivial group 2
homomorphism, and f2 : Z/2Z × Z/2Z → GL2(R); (0, 0) 7→ I2, (1, 0) 7→