Solutions of Assignment 10 Basic Algebra I

November 25, 2004

Solution of the problem 1. Let |a| = m, |bab^{−1}| = n. Since

(bab^{−1})^{m} = (bab^{−1})(bab^{−1}) ⋯ (bab^{−1})
= ba^{m}b^{−1}
= b1b^{−1}
= 1,

we have n ≤ m. Conversely, since

a^{n} = b^{−1}ba^{n}b^{−1}b
= b^{−1}(bab^{−1})^{n}b
= b^{−1}1b
= 1,

we have m ≤ n. Thus m = n.

Solution of the problem 2. Recall A non-empty subset H of a group G is a subgroup iff it satisfies the following property:

∀h_{1}, h_{2} ∈ H ⇒ h_{1}h_{2}^{−1} ∈ H. (⋆)

Now back to our problem, we check that H = H_{1} ∩ H_{2} satisfies (⋆): Take h_{1}, h_{2} ∈ H. So, h_{1}, h_{2} ∈ H_{1}; h_{1}, h_{2} ∈ H_{2}. Since both H_{1} and H_{2} are assumed to be subgroups of G, then (⋆) tells us that

h_{1}h_{2}^{−1} ∈ H_{1}, h_{1}h_{2}^{−1} ∈ H_{2}.

Therefore h_{1}h_{2}^{−1} ∈ H.

For the union, we shall prove the following:

H_{1} ∪ H_{2} is a subgroup of G iff either H_{1} ⊆ H_{2} or H_{2} ⊆ H_{1}.

Proof Sufficiency is clear. So, suppose that H_{1} ∪H_{2} is a subgroup of G, and, on the contrary, assume that H_{1} ⊄ H_{2} and that H_{2} ⊄ H_{1}. These in return imply that

∃h_{1} ∈ H_{1}, s.t. h_{1} ∉ H_{2}; ∃h_{2} ∈ H_{2}, s.t. h_{2} ∉ H_{1}.

Since h_{1}h_{2} ∈ H_{1} ∪ H_{2} (why?), then we would have either h_{1}h_{2} ∈ H_{1} or h_{1}h_{2} ∈ H_{2}, and both are impossible (why?). Done.

Solution of the problem 3. Let |a| = m. By Lagrange’s theorem, m | n. So,

a^{n} = (a^{m})^{\frac{n}{m}} = 1^{\frac{n}{m}} = 1.

For the second part, if a ≡ 0 (mod p), then it is evident that

p | a(a^{p−1} − 1) = a^{p} − a.

And if a ≢ 0 (mod p), then a ∈ ℤ_{p}^{×}, and since ℤ_{p}^{×} is a group of order p − 1, by what we proved above, a^{p−1} = 1 (in ℤ_{p}^{×}), so

p | a(a^{p−1} − 1) = a^{p} − a.

Solution of the problem 4. We verify that Z(S) satisfies (⋆) in the solution of problem 2: Let a, b ∈ Z(S). So, as = sa, bs = sb for s ∈ S. First note that sb^{−1} = b^{−1}bsb^{−1} = b^{−1}sbb^{−1} = b^{−1}s. Therefore

(ab^{−1})s = ab^{−1}s = asb^{−1} = sab^{−1} = s(ab^{−1}),

hence ab^{−1} ∈ Z(S).

Solution of the problem 5. Define ϕ : G_{1} ⟶ G_{2}, ϕ(x) = ln(x). ϕ is clearly bijective. Also note that

ϕ(xy) = ln(xy) = ln(x) + ln(y) = ϕ(x) + ϕ(y).

So, ϕ is homomorphism, hence an isomorphism.

Solution of the problem 6. Let |a| = m, |f(a)| = n. Since f is a homomorphism, we have

f(a)^{m} = f(a^{m}) = f(1_{G_{1}}) = 1_{G_{2}}.

So, n ≤ m. If f is also injective, we have

f(a^{n}) = f(a)^{n} = 1_{G_{2}} = f(1_{G_{1}}).

So, a^{n} = 1_{G_{1}}, since f is injective. Thus m ≤ n and we are done.

Solution of the problem 7. Note that:

(i) G is closed under multiplication (check this);

(ii) G contains the identity element 1 (clear);

(iii) G contains the inverse of all its elements:

(±1)^{−1} = ±1, (±i)^{−1} = ∓i, (±j)^{−1} = ∓j, (±k)^{−1} = ∓k.

Hence G is a (sub)group of the multiplicative group of non-zero elements of H.

For the second part, enough to see that the dihedral group D_{4} has two elements of order 4, namely r_{1} and r_{3}, whereas in the group G above, there are six elements of order 4, namely ±i, ± j, ± k. So, G ≅ D̸_{4}.

Extra Credit

Solution of the problem 9. Let V = ℤ_{2} × ℤ_{2} = {(0, 0), (1, 0), (0, 1), (1, 1)} = {o, e_{1}, e_{2}, e_{3}}, where o = (0, 0), e_{1} = (1, 0), e_{2} = (0, 1), e_{3} = (1, 1). We will view V as a vector space of dimension 2 over the field ℤ_{2}. Fix the basis {e_{1}, e_{2}} for V. Now each matrix

\binom{a b}{c d} ∈ GL_{2}(ℤ_{2})

may be viewed as a bijective linear transformation from V into itself (by mul-

tiplication from left to e_{i}’s). Each \binom{a b}{c d} permutes e_{1}, e_{2}, e_{3}. For example

\binom{1 1}{0 1}e_{1} = \binom{1 1}{0 1}\binom{1}{0} = \binom{1}{0} = e_{1},

\binom{1 1}{0 1}e_{2} = \binom{1 1}{0 1}\binom{0}{1} = \binom{1}{1} = e_{3},

\binom{1 1}{0 1}e_{3} = \binom{1 1}{0 1}\binom{1}{1} = \binom{0}{1} = e_{2}.

So, \binom{1 1}{0 1} can be corresponded to the permutation σ = \binom{1 2 3}{1 3 2}.

In general, we can define a well-defined map, ψ say, from the group GL_{2}(ℤ_{2}) into the group S_{3}:

\binom{a b}{c d} ↦ σ = \binom{1 2 3}{σ(1)σ(2)σ(3)},

where

\binom{a b}{c d}e_{i} = e_{σ(i)} (1 ≤ i ≤ 3).

ψ is clearly an injective group homomorphism (check this). On the other hand, since |S_{3}| = |GL_{2}(ℤ_{2})| = 6, we conclude that ψ is also onto, hence an isomorphism.

Solution of the problem 10. Let G be a group, and let a, b ∈ G. b is said to be conjugate to a if b = gag^{−1} for some g ∈ G. Notice that

(i) Every a is conjugate to itself: a = 1a1^{−1};

(ii) If b is conjugate to a, then a is also conjugate to b:

b = gag^{−1} ⇒ a = g^{−1}b(g^{−1})^{−1};

(iii) If b is conjugate to a, and if c is conjugate to b, then c is also conjugate to a:

b = g_{1}ag_{1}^{−1}, c = g_{2}bg_{2}^{−1} ⇒ c = (g_{2}g_{1})a(g_{2}g_{1})^{−1}.

So, conjugacy is an equivalence relation in G. Denote the conjugacy class of a ∈ G by cl[a]:

cl[a] := {gag^{−1} : g ∈ G}.

Now suppose that N ⊳ G, i.e., N is a normal subgroup of G. Given any a ∈ N, it is obvious that cl[a] ⊆ N (why?). Thus N is the disjoint union of the conjugacy classes of its elements. Conversely, if a subgroup of G is a union of some conjugacy classes in G, that subgroup is clearly normal. So, one way to find all normal subgroups of G is to look at those unions of conjugacy classes in G which constitute a subgroup.

To determine the conjugacy classes in the symmetric group S_{n}, we will exploit the following useful fact:

Permutations α, β ∈ S_{n} are conjugate iff the have the same cyclic structure, i.e., iff their complete factorization into disjoint cycles have the same number of r-cycles for each r.

Example Let

α = (2 3 1)(4 5)(6);
β = (5 6 2)(3 1)(4);
γ = (2 3 1)(4 5 6).

α and β are conjugate, since they have the same cyclic structure. In fact the

permutation δ = \binom{1 2 3 4 5 6}{2 5 6 3 1 4} does what we want: δαδ^{−1} = β (check this). In

complete contrast, α and γ are not conjugate, because they don’t have the same cyclic structure.

Using the above fact, now listing the set of all conjugacy classes in S_{4} is an easy(!) task:

C_{1} = {(1)};
C_{2} = {(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)};
C_{3} = {(1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)};
C_{4} = {(1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)};
C_{5} = {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

Examining all the possibilities, one can find all the normal subgroups of S_{4}:

{1} = C_{1}; V = C_{1} ∪ C_{5}; A_{4} = C_{1} ∪ C_{3} ∪ C_{5}; S_{4} = C_{1} ∪ C_{2} ∪ C_{3} ∪ C_{4} ∪ C_{5}.

As for S_{5}, the following is the complete list of all conjugacy classes:

cl[(1)]; cl[(1 2)]; cl[(1 2 3)]; cl[(1 2 3 4)];

cl[(1 2 3 4 5)]; cl[(1 2)(3 4)]; cl[(1 2 3)(4 5)].

And finally, one can find all normal subgroups of S_{5}. Here you are:

{1}; A_{5}; S_{5}.

Conclusion A_{5} is the only proper non-trivial normal subgroup of S_{5}. In fact, this holds for any n ≠ 4.