Download Math. 5363, exam 1, solutions 1. Prove that every finitely generated

Math. 5363, exam 1, solutions
1. Prove that every finitely generated abelian group is the homomorphic image of
a free finitely generated abelian group.
Let G be a finitely generated abelian group and let {g1 , g2 , ..., gn } ⊆ G be a set of
generators. Let {x1 , x2 , .., xn } be a Z-basis of Zn . Then
f : Zn 3 a1 x1 + a2 x2 + ... + an xn → a1 g1 + a2 g2 + ... + an gn ∈ G
is a well defined surjective group homomorphism and G = f (Zn ).
2. Describe all the abelian groups of order 44, up to an isomorphism.
The classification of the finitely generated abelian groups implies that a group of
order 44 = 4 × 11 is isomorphic either to C4 × C11 or to C2 × C2 × C11 .
3. Show that the group of the characters of the additive Z is isomorphic to the
multiplicative group T of the complex numbers of absolute value 1.
Define the following map
(3.1)
Char(Z) 3 χ → χ(1) ∈ T.
Since, for two characters χ and χ0
(χχ0 )(1) = χ(1)χ0 (1)
the map (3.1) is a group homomorphism.
Suppose χ is in the kernel of the homomorphism (3.1). Then
χ(1) = 1
Then χ(n) = χ(1)n = z n for all n ∈ Z. Thus χ is a trivial character. Therefore the
map (3.1) is injective.
1
2
Given z ∈ T define
χ(n) = z n
(n ∈ Z).
Then, (3.1) maps this character to z. Thus (3.1) is surjective.
4. State Sylow’s Theorem.
Look it up in the book.
5. Are the Sylow 2-groups in S4 abelian or not. Why?
Let K ⊆ S4 be the subgroup generated by the cycle (1234). This group has 4
elements. Since |S4 | = 3 × 8, there is a Sylow 2-subgroup H ⊆ S4 containing K.
Notice that for any permutation σ ∈ S4 ,
σ(1234)σ −1 = (σ(1)σ(2)σ(3)σ(4)).
Hence, σ commutes with the cycle iff
(σ(1)σ(2)σ(3)σ(4)) = (1234),
which means that σ is a power of (1234). Thus there is no commutative extension
of K in S4 . In particular H is not commutative.
6. Suppose G is a finite group, H is a subgroup, and a ∈ G is an element with
an ∈ H for some integer n with GCD(n, |G|) = 1. Prove that a ∈ H.
Since GCD(n, |G|) = 1, there are integers x, y such that xn + y|G| = 1. Hence,
a = a1 = axn ay|G| = axn (a|G| )y = axn = (an )x ∈ H.
7. Let G be a group of order 2pq, where p and q are primes with 2 < p < q. Prove
that if q + 1 6= 2p then the Sylow q-subgroup is normal.
3
The number of Sylow q subgroups of G is equal to qk + 1, where k is a non-negative
integer. Furthermore, qk +1 divides 2p. Since 2 < p < q, we must have k = 0. Thus
there is only one Sylow q subgroup. Since all Sylow q subgroups are conjugate, this
unique subgroup is normal.
8. Let G be a non-abelian group of order 6. Prove that G is isomorphic to S3 .
Let G be a non-abelian group of order 6. Since G is not abelian, it does not contain
any element of order 6. Also, it can’t happen that every element other than 1 is of
order 2. Therefore, there is element a ∈ G of order 3.
This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In particular, H is a normal subgroup.
Since |G| = 2 × 3, there is a Sylow subgroup of G of order 2. In other words,
there is an element b ∈ G of order 2. Since H does not contain any elements of
order 2, we see that b ∈ G and b ∈
/ H.
The conjugation by b preserves H, but does not commute with a. Hence, bab−1 =
a2 . Thus G = {1, a, a2 , b, ab, a2 b} and the following relation holds
(8.1)
ba = a2 b.
4
The relation (8.1) determines the multiplication table for G. (Explicitly,
a · a = a2 ,
a · a2 = a3 = 1,
a · b = ab,
a · ab = a2 b,
a · a2 b = b,
a2 · a = 1,
a2 · a2 = a,
a2 · b = a2 b,
a2 · ab = b,
a2 · a2 b = ab,
b · a = a2 b,
b · a2 = a2 ba = a4 b = ab,
b · b = 1,
b · ab = a2 b2 = a2 ,
b · a2 b = a2 bab = a4 b2 = a,
ab · a = aa2 b = b,
ab · a2 = aa2 ba = ba = a2 b,
ab · b = a,
ab · ab = aa2 bb = 1,
ab · a2 b = abba = a2 ,
a2 b · a = ab,
a2 b · a2 = b,
a2 b · b = a2 ,
a2 b · ab = a,
a2 b · a2 b = 1.)
Thus up to a group isomorphism, there is only one non-abelian group of order 6.
In particular G is isomorphic to S3 .