Exercises on Compatible relations and normal subgroups

1. Determine the quotient group of S₃ with respect to its unique non-trivial normal subgroup.

2. Verify that in (GL_n(ℝ), ⋅) the subgroup (SL_n(ℝ), ⋅) is a normal subgroup. Which group is isomorphic to?

3. Verify that the subgroup (SO₂(ℝ), ⋅) made by orthogonal matrix with unitary determinant is a normal subgroup of the orthogonal matrices (O₂(ℝ), ⋅). How many elements Does the quotient group has?

   The orthogonal group O_n(ℝ) is the subset of M_n(ℝ) given by O_n(R) = {M ∈ M_n(R), M^tM = MM^t = I_n} where M^t denotes the transpose of a matrix M. We recall that for any matrix M,M and M^t have the same determinant.

4. Find all normal subgroups of S₄.

5. Prove that the alternating subgroup A_n is normal in S_n and analyze the quotient group.

6. Let H and K two subgroups of a group G and let H ⊲̲ K. Give an example to show that it is not generally true that H ⊲̲ G.

7. Given the elements of the alternating group A₄ of S₄. Find the subgroup of H of A₄ generated by (124). Find all distinc left and right coset of H in A₄. Determine if H is a normal subgroup of A₄.

Solutions

1. We have seen that A₃ is a normal subhgroup of S₃; therefore S₃/A₃ is a quotient group. If H = A₃ = {e, (123), (132)}, the elements of this group are the cosets H and H(12) = {(12),(13),(23)} (Notice how (12)(13)⁻¹ ∈ A₃). By setting 𝓑 = H(12) we have S₃/A₃ = {A₃, 𝓑} is a cyclic group of order 2. The multiplicative relations are

   A₃A₃ = A₃,   𝓑𝓑 = A₃,   𝓑A₃ = 𝓑,   𝓑A₃ = A₃𝓑 = 𝓑

   A₃ is the identity. Therefore S₃/A₃ ≈ S₂.

   A₃ → e,   𝓑 → (12).

2. First we prove it is indeed a subgroup. Binet's theorem states that det(AB) = det(A)det(B), for all matrices A, B ∈ M_n(ℝ), where det(A) denotes the determinant of A. Let

   (SL_n(ℝ), ⋅) = {A ∈ M_n(ℝ) | det A = 1}

   a nonempty set. If A, B ∈ SL_n(ℝ), then det(AB) = det(A)det(B) = 1, from which it follows that AB ∈ SL_n(ℝ). Furthermore,

   1 = det(I) = det(AA⁻¹) = det(A)det(A⁻¹),

   and therefore det(A⁻¹) = 1, so that A⁻¹ ∈ SL_n(ℝ) also. Consequently, SL_n(ℝ) satisfies both subgroups conditions, and hence it is a subgroup of GL_n(ℝ).

   Now we show that this subgroup is normal. Indeed, let A ∈ SL_n(ℝ) and B ∈ GL_n(ℝ) . Then

   det(B⁻¹ AB) = det(B⁻¹)det(A)det(B) = 1/det(B) ⋅ det(A) ⋅ det(B) = det(A) = 1,

   which implies that B⁻¹AB ∈ SL_n(ℝ) and Propositions 7.11.7 shows that SL_n(ℝ) is a normal subgroup of GL_n(ℝ).

   The quotient group is made by all cosets of matrices which are equivalent modulo SL_n(ℝ) i.e: A ρ B ⇐⇒ AB⁻¹ ∈ in SL_n(ℝ), that is those having the same determinant since det(AB⁻¹) = det A/det B = 1. GL_n(ℝ)/SL_n(ℝ) is then isomorphic to ℝ \{0}.   ■

3. Let A ∈ (SO₂(ℝ), ⋅) and B ∈ (O₂(ℝ), ⋅). Then

   det(B⁻¹ AB) = det(B⁻¹)det(A)det(B) = 1/det(B) ⋅ det(A) ⋅ det(B) = det(A) = 1,

   which implies that B⁻¹ AB ∈ SL_n(ℝ) and Propositions 7.11.7 shows that (SO₂(ℝ), ⋅) is a normal subgroup of O₂(ℝ), ⋅). The quotient group is made by all cosets of matrices which are equivalent modulo SO₂(ℝ) i.e.: A ρ B ⇐⇒ AB⁻¹ ∈ in SO₂(ℝ), that is those having the same determinant since det(AB⁻¹) = det A/det B = 1, for A and B orthogonal. Since for an orthogonal matrix, Q, det Q = ±1 we have

   O₂(Rℝ)/SO₂(ℝ) = { {matrices with det = 1}, {matrices with det = -1}}

   Thus the quotient group is isomorphic to {±1}.

   O₂(ℝ) is the set of rotations and reflections of the plane (or of the unit circle if you prefer). Its subgroup SO₂(R) is the set of rotations. Both groups are infinite (indeed uncountable), but we can compute the index: (O_n(ℝ): SO_n(ℝ)) = 2.   ■

4. By proposition N is a union of conjugacy classes of S₄. In S₄ the conjugacy class of

   (id)   has 1 element.
   (12)   has 6 elements.
   (123)   has 8 elements.
   (1234)   has 6 elements.
   (12)(34)   has 3 elements.

   By the Lagrange theorem |N| divides 24. The only possibilities, are 1, 1 + 3, 1 + 3 + 8, 1 + 6 + 3 + 8 + 6. Note that 1 + 6 + 3 + 8 + 6 = 24 = |S₄|, as should be the case since conjugacy classes of a group G form a partition of G. So the only possible orders for non-trivial proper normal subgroups are 4 and 12. The 1 + 3 case gives the subgroup

   {(id), (12)(34), (13)(24), (14)(23)}

   To see that is a normal subgroup, it suffices to check that it is a subgroup from the fact

   (12)(34) ⋅ (13)(24) = (14)(23), etc.

   1 + 3 + 8, gives

   {id, (123), (132), (234), (243), (134), (143), (124), (142), (12)(34), (14)(23)}.  ■

5. A_n is the subgroup of even permutation in S_n. The subgroup A_n is a normal subgroup of S_n, since the conjugation h → g ⋅ h ⋅ g⁻¹ preserves the parity of h. By Lagrange theorem |S_n| = n! and |A_n|= n!/2 so [S_n: A_n] = 2, that is |S_n/A_n|= 2. Since the identity element of the quotient group is the class of even permutations A_n, the other class is the one containing odd permutations of S_n that is S_n/A_n = {A_n, σA_n, where σ is an odd permutation of S_n

   For a double check: Like all groups of two elements, S_n/A_n must be cyclic: the nonidentity element must have square equal to the identity. The definition of coset multiplication shows:

   A_nσ ⋅ A_nσ = A_nσ² = A_n ⋅ id = A_n

   where σ is an odd permutation, or put it in another way σ² A_n = A_n, because any σ² (for any p in any S_n at all) must be an even permutation, so your σ² is in that A_n.  ■

6. Take a look at Cayley Table for D₄. You should find H = {e, rs}, K = {e, r², rs , r³s} in D₄. We have H ⊲ K but H ⋪ G.

7. Every 3-cycle (abc) in A₄ has order 3. H = ⟨ (124)⟩ {(124), (124)² = (142) = (124)⁻¹, (124)³ = id}. It's not normal since it's not the union of conjugacy classes.

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