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Final Exam (Solutions)
May 11, 2011
1. Classify the group G = (Z × Z)/〈(5, 6)〉 according to the Fundamental Theorem of Theory of
Finitely Generated Abelian Groups.
Let π : Z×Z → G be the quotient map and ι : 〈(1, 1)〉 → Z×Z be the inclusion map. We will show that φ = π ◦ ι is an isomorphism which means that G is infinite cyclic. Indeed, 〈(1, 1), (5, 6)〉 contains both (1, 0) and (0, 1) :
(1, 0) = 6 · (1, 1) − (5, 6) and (0, 1) = − 5 · (1, 1) + (5, 6)
which generate Z × Z , and thus 〈(1, 1), (5, 6)〉 = Z × Z_. It follows that φ is surjective. It is obviously injective, since_ 〈(1, 1)〉 ∩ 〈(5, 6)〉 = {(0, 0)}.
2. Show that the rings 2Z and 3Z are not isomorphic.
Any ring isomorphism is an isomorphism of the corresponding additive groups. Both 2 Z and 3 Z are infinite cyclic, and an isomorphism between 2 Z and 3 Z must take 2 to a generator of 3 Z , i.e. to 3 or − 3_. In partic- ular,_ 4 = 2 + 2 would be sent to either 6 = 3 + 3 , or − 6 = − 3 + (−3) instead of 9 = (±3)^2_. Thus, none of the two isomorphisms of the additive groups is a homomorphism of multiplicative semigroups._
3. Let
and β =
Find σ ∈ S 8 such that σ ◦ α = β ◦ σ.
One has α = (2 5 6)(3 8 4 7) and β = (4 7 8)(1 2 5 6). It follows that if we put
σ =
then σ ◦ α ◦ σ −^1 = β.
4. Let α be the permutation defined in formula (1) above. Find α^2011.
Permutation α is the product of disjoint cycles λ = (2 5 6) and μ = (3 8 4 7). Hence
α^2011 = λ^2011 μ^2011 = λμ −^1 =
in view of the fact that 2011 = 1 mod 3 and 2011 = − 1 mod 4_._
5. Find the order of the permutation σ = (3 11 5)(10 5 4 3 2 11 6)(3 5 11) in S 11.
Note that (3 5 11) = (3 11 5)−^1 , so σ is a conjugate of a cycle of length 7 , thus is a cycle iteslf and its order equals its length, i.e. 7_._
6. Let G be a group. We say that G -sets X and X ′^ are isomorphic if there exists a bijection
f : X → X ′^ such that f ( g · x ) = g · f ( x ) for any g ∈ G and x ∈ X. Prove that the orbit G x of any
element x ∈ X is isomorphic to G / Gx.
The obvious map φ : G → X , given by g 7 → g · x , has the desired property and is surjective. One has g · x = g ′^ · x precisely if g −^1 g ′^ ∈ Gx , i.e., φ ( g ) depends only on the coset [ g ] = gGx , and the induced map f : G / Gx → G x , where f ([ g ]) = g · x , is a desired isomorphism.
7. Let X be a G -set and O ⊆ X be any orbit. Prove that |O| divides | G | if G is finite.
The assertion of Problem 6 implies that, for any x ∈ O , one has |O| = | G / Gx | = | G : Gx | , and the index of any
subgroup of G divides the order of G if G is finite (Lagrange’s Theorem).
8. Prove that any subgroup H ⊂ G of index 2 is normal.
G \ H is the union of left cosets distinct from H and also the union of right cosets distinct from H. Since index of H in G is 2 , G \ H consists of a single left, and of a single right coset. Thus, every left coset is a right coset and vice-versa, i.e., H is normal in G.
9. Prove that any action of a group G of order 11 on a set X with 2011 elements must have at least
9 fixed points.
Set X is the union of disjoint orbits. By Problem 7, |O| divides | G | = 11 , so equals 1 or 11. Orbits of cardinality
1 correspond to fixed points. Denote by m the number of orbits having 11 elements. It follows that 2011 = | X | = 11 m + | FixG ( X )| and, since 11 m ≤ 2011 and the largest multiple of 11 less or equal 2011 is 2002 , we conclude that | FixG ( X )| ≥ 2011 − 2002 = 9_._
