Math 571 Homework 2 Kevin Childers
Proof. Assume g ∈ G. Then for any z ∈ Z(G), then zg = gz by definition of the center.
So g ∈ CG(Z(G)). The other direction is trivial. Therefore, CG(Z(G)) = G.
Since CG(Z(G) ≤ NG(Z(G)), NG(Z(G)) = G.
4 For G = S3, there are three types of elements. Obviously CG(1) = G since everything commutes with 1. For a transposition σ, there are at least two elements in CG(σ), 1 and σ. But this group must have order divisible by 2 and dividing 6. But any 3-cycle will not commute with any transposition so it cannot be the whole group. Therefore those are the only two elements. Similarly with any 3-cycle τ, the other 3-cycle will be 2 τ , so CG(τ) has at least three elements. But it can’t have all 6 because transpositions will not commute. Therefore that is the whole subgroup.
For G = D8,
CG(1) = G ∼ CG(r) = hri = C4 2 CG(r ) = G 3 ∼ CG(r ) = hri = C4 2 ∼ CG(s) = hr , si = V4 2 ∼ CG(sr) = hsr, r i = V4 2 2 ∼ CG(sr ) = hr , si = V4 3 2 ∼ CG(sr ) = hsr, r i = V4.
And for G = Q8,
CG(1) = G
CG(−1) = G ∼ CG(i) = hii = C4 ∼ CG(−i) = hii = C4 ∼ CG(j) = hji = C4 ∼ CG(−j) = hji = C4 ∼ CG(k) = hki = C4 ∼ CG(−k) = hki = C4.
5 (a) G = S3, A = {1, (1 2 3), (1 3 2)}. CG(A) = A as above. But for any transposition τ and 3-cycle σ, τστ will be a 3-cycle, so τAτ ∈ A, and NG(A) = G. 2 2 3 2 (b) G = D8 and A = {1, s, r , sr }. CG(A) = A also from above. But rsr = rrs = sr , 2 3 3 rsr r = rsr = rr s = s so r ∈ NG(A) but since the order of this subgroup must divide 8, NG(A) = G.
1 2 3 4 4 (c) G = D10, A = {1, r, r , r , r }. Since sr = r s, s 6∈ CG(A). Since the order of this a −a −a subgroup divides 10, CG(A) = A. However, sr s = ssr = r ∈ A, so s ∈ NG(A), and once again the order of this subgroup divides 10, so NG(A) = 10.
6 Let H ≤ G.
(a) H ≤ NG(H).
−1 Proof. Obviously for any h, k ∈ H, hkh ∈ H, therefore H ≤ NG(H).
2 Take A = {r, s} ⊆ D8. Then NG(A) = {1, r }. A 6⊆ NG(A). (b) H ≤ CG(H) if and only if H is abelian.
Proof. Assume H ≤ CG(H). Then for any h, k ∈ H, hk = kh, therefore H is abelian. Assume H is abelian. Then for any h, k ∈ H, hk = kh, therefore h ∈ CG(H). So in fact H ≤ CG(H).
7 Let n ∈ Z with n ≥ 3, then (a) Z(D2n) = 1 if n is odd, and k (b) Z(D2n) = {1, r } if n = 2k.
i Proof. (a) First We show that no sr ∈ Z(D2n).
(sri)r(sri)−1 = srirr−is = srs = r−1.
So if any non-trivial element is in the center, it is of the form ri. Also we need for any a sr ∈ D2n that ri = (sra)ri(sra)−1 = srarir−is = sris = r−i or in other words, i ≡ −i (mod n). If n is odd, this never happens. If n = 2k, this only happens when i = k.
10 Let H ≤ G with |H| = 2. Then NG(H) = CG(H).
Proof. Say that H = {e, h} where e is the identity. We already know that CG(H) ≤ −1 −1 −1 NG(H). Let g ∈ NG(H). Then gHg = H. So {geg , ghg } = {e, h}. But since −1 −1 −1 geg = gg = e we must have ghg = h which implies that g ∈ CG(H). So NG(H) = CG(H).
Supposing NG(H) = G, this implies that CG(H) = G. But then all elements of H commute with any element of G, so H ≤ Z(G).
12 Let R = Z[x1, x2, x3, x4] and let S4 act upon R as described, and let
5 7 3 6 3 23 p(x1, x2, x3, x4) = 12x1x2x4 − 18x2x3 + 11x1x2x3x4 .
2 (a) Let σ = (1 2 3 4) and τ = (1 2 3). Then σ ◦ τ = (1 3 2 4) and τ ◦ σ = (1 3 4 2).
5 7 3 6 3 23 σ · p = 12x2x3x1 − 18x3x4 + 11x2x3x4x1 5 7 3 6 3 23 τ · (σ · p) = 12x2x3x4 − 18x3x1 + 11x2x3x1x4 5 7 3 6 3 23 (τ ◦ σ) · p = 12x3x4x1 − 18x4x2 + 11x3x4x2x1 5 7 3 6 3 23 (σ ◦ τ) · p = 12x3x1x2 − 18x1x4 + 11x3x1x4x2 (b) The action forms a left group action.
Proof. Given p(x1, x2, x3, x4) ∈ R and σ, τ ∈ S4, we have
(σ ◦ τ) · p = p(xσ(τ(1)), xσ(τ(2)), xσ(τ(3)), xσ(τ(4)))
= σ · p(xτ(1), xτ(2), xτ(3), xτ(4)) = σ · (τ · p(x1, x2, x3, x4)).
And for the identity element 1 ∈ S4, 1 · p = p(x1, x2, x3, x4), therefore the action is a group action.
(c) Let σ ∈ Stab(x4). Then σ(4) = 4, which happens for the permutations in the set
{1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} = S3.
(d) If σ ∈ Stab(x1 + x2), then {σ(1), σ(2)} = {1, 2}. This happens exactly for the elements of the set
{1, (1 2), (1 2)(3 4), (3 4)} = h(1 2), (3 4)i
and all elements of this set commute.
(e) If σ ∈ Stab(x1x2 + x3x4), then
{{σ(1), σ(2)}, {σ(3), σ(4)}} = {{1, 2}, {3, 4}}.
This will happen only for elements of the set
{(1), (1 2), (3 4), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 4 2 3), (1 3 2 4)} = h(1 4 2 3), (1 4)(2 3)i
which is isomorphic to D8 by labelling the corners of the square 1, 4, 2, 3 going counter clockwise.
(f) If σ ∈ Stab((x1 + x2)(x3 + x4)) then once again we have the following equivalence of sets: {{σ(1), σ(2)}, {σ(3), σ(4)}} = {{1, 2}, {3, 4}}. Therefore such elements will form the same group as in part (e).
14 Let H(F ) be the Heisenberg group over a field F . If a matrix A ∈ Z(H(F )), then for any B ∈ H(F ), letting
1 a b 1 e f A = 1 c ,B = 1 g 1 1
3 we need AB = BA. This yields the equations
a + e = e + a c + g = g + c b + ce + f = f + ag + b.
The first two are always true, and the third implies that ce = ag for all e, g ∈ F which only happens if a = c = 0. Therefore Z(H(F )) is the set of all matrices of the form
1 0 a 1 0 1 with a ∈ F . And the function from (F, +) → (Z(H(F )), ·) defined by
1 0 a a 7→ 1 0 1 is the obvious isomorphism of groups since
1 0 a 1 0 b 1 0 a + b 1 0 1 0 = 1 0 . 1 1 1
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