Download MTH 818 Algebra I F12 Homework 2 / Solutions Lemma A. Let G be

MTH 818
Algebra I
F12
Homework 2 / Solutions
Lemma A. Let G be a group and H ≤ G. Let G/op H be the set of right cosets of H in G. Then the map
τ ∶ G/H → G/op H, A → A−1
is a well defined bijection.
Proof. Let g ∈ G. The (gH)−1 = H −1 g−1 = Hg−1 and so τ(gH) ∈ G/op H. Thus τ is well defined. The map
σ ∶ G/op H → G/H, A → A−1
is an inverse of τ and so τ is a bijection.
# 1. Let G be a group and H ≤ G with ∣G/H∣ = 2. Show that H ⊴ G.
Proof. Note that H is one of the two cosets G. Let A be the other. Since G is the disjoint union of it cosets,
G = H ⊍ A and so A = G ∖ H. Thus the cosets of H are H and G ∖ H. By Lemma A ∣G/op H∣ = ∣G/H∣ = 2. It
follows that H and H ∖ G are also the right coset of H in G. So H fulfills one of conditions in Lemma 1.5.3
(namely part(c)) and thus H ⊴ G.
Seems I never proved the Cancellation Law in class:
Lemma B. Let G be a group and a, b, c ∈ G. Then
(a) (ac)c−1 = a and c−1 (ca) = a.
(b) ac = bc
⇐⇒
a=b
(c) ac = a
⇐⇒
c=1
(d) ac = 1
⇐⇒
c = a−1
−1
⇐⇒
⇐⇒
⇐⇒
ca = cb.
ca = a.
ca = 1.
−1
Proof. (a) (ac)c = a(cc ) = a1 = a. This result applied to the opposite group gives the second statement.
(b) Suppose ac = bc. Then (ac)c−1 = (bc)c−1 and by (a) a = b. Clearly a = b implies ac = bc. Thus
ac = bc ⇐⇒ a = b. Thus result applied to the opposite group gives ca = cb ⇐⇒ a = b.
(c) Since ac = a ⇐⇒ ac = a1, (c) follows for (b).
(d) Since ac = 1 ⇐⇒ ac = aa−1 , (d) follows from (b).
# 2. Let φ ∶ G → H be a homomorphism of groups. Prove that
(a) φ(1G ) = 1H .
(b) φ(a−1 ) = φ(a)−1 for all a ∈ G.
(c) If I is a subgroup of G, then φ(I) is a subgroup of H.
(d) If J is a subgroup of H, then φ−1 (J) is a subgroup of G.
(e) If J is a normal subgroup of H, then φ−1 (J) is a normal subgroup of G.
(f) If I is a normal subgroup of G and φ is onto, then φ(I) is a normal subgroup of H.
Proof. (a) φ(1G ) = φ(1G 1G ) = φ(1G )φ(1G ) and so Lemma B(c) gives φ(1G ) = 1H .
(a)
(b) φ(a)φ(a−1 ) = φ(aa−1 ) = φ(1G ) = 1H and so Lemma B(d) gives φ(a1 ) = φ(a)−1 .
(c) Since I is a subgroup of G, 1G ∈ I. Thus 1H = φ(1G ) ∈ φ(I). Let x, y ∈ φ(I). Then x = φ(i) and
y = φ(k) for some i, k ∈ I. Since I is a subgroup of G, ik ∈ I and i−1 ∈ I. Thus xy = φ(i)φ( j) = φ(ik) ∈ φ(I)
and x−1 = φ(i)−1 = φ(i−1 ) ∈ φ(I). Thus I is a subgroup of H.
(d) Let i, k ∈ φ−1 (J). Then φ(i), φ(k) ∈ J. Since J is a subgroup of H we have 1H ∈ J, φ(i)φ(k) ∈ J and
φ(i)−1 ∈ J. Hence φ(1G ) ∈ J, φ(ik) ∈ J and φ(i−1 ) ∈ J. It follows that 1H , ik and i−1 all are in φ−1 (J) and so
φ−1 (J) is a subgroup of G.
For later use we prove:
1○ .
Let a, b ∈ G. Then φ (ab) = φ(a)φ(b).
φ(ab) = φ(aba−1 ) = φ(ab)φ(a−1 ) = φ(a)φ(b)φ(a)−1 = φ(a)φ(b)
(b)
(e) By (d), φ−1 (J) is a subgroup of G.
Let i ∈ φ−1 (J) and g ∈ G. Then φ(i) ∈ J and since J ⊴ G, φ(g)φ(i) ∈ J. Thus by (1○ ) φ(gi) ∈ J and so
g
i ∈ φ−1 (J). Thus φ−1 (J) ⊴ G.
(f) By (c), φ(I) is a subgroup of G. Let j ∈ φ(I) and h ∈ H. Then j = φ(i) for some i ∈ I and since φ is
onto, h = φ(g) for some g ∈ G. Since I ⊴ G, gi ∈ I. Hence using (1○ )
h
j = φ(g)φ(i) = φ(gi) ∈ φ(I)
Thus φ(I) ⊴ G.
# 3. Let G be a group, H ⊆ G, K ⊆ G and N ≤ G. If H ⊆ N then
(HK) ∩ N = H(K ∩ N).
Proof. Note that H(K ∩ N) ⊆ NN ⊆ N and H(K ∩ N) ⊆ HK. Thus H(K ∩ N) ⊆ (HK) ∩ N.
If g ∈ HK ∩ N, then g ∈ N and g = hk for some h ∈ H, k ∈ K. Then k = h−1 g. Note that h ∈ H ⊆ N and
g ∈ N. Since N is a subgroup of G, we get k = h−1 g ∈ N and so k ∈ K ∩ N. Thus g = hk ∈ H(K ∩ N) and
HK ∩ N ⊆ H(K ∩ N).
# 4. Let G be a group and H ≤ K ≤ G. Let T ∈ G/K.
(a) Show that
T = ⋃ S.
S ∈T /H
(b) Let g ∈ T (and so T = gK). Show that the map
β ∶ T /H → K/H, S → g−1 S
is well-defined bijection.
(c) Show that ∣G/H∣ = ∣G/K∣ ⋅ ∣K/H∣.
Proof. (a) Recall from Example 1.6.16(2) that T is an orbit of K on G with respect to the action of K on G by
right multiplication. In particular, K acts on T (by right multiplication) and since H ≤ K, H acts on T . The
orbits of H on T are the cosets tH, t ∈ T and so T /H is the set of orbits of H on T . Hence
T= ⊍ S
S ∈T /H
(b) By Example 1.6.16(5), G acts on G/H by left multiplication. Since S ∈ G/H we conclude that also
g−1 S ∈ G/H. Since both g and S are contained in T and T is a coset of K, g−1 S ⊆ K. Thus g−1 S ∈ K/H and
β is well defined.
Consider the map
γ ∶ K/H → T /H, U → gU
Since U is a coset of H, so is gU. Also gU ⊆ gK = T and so gU ∈ T /H. So γ is well defined. Since
g−1 (gU) = U and g(g−1 S ) = S . γ is the inverse of β and so (b) is proved.
(c) Let S ∈ G/H. Then there exists a unique coset T of K with T ∩ S ≠ ∅, namely S K (indeed if s ∈ T ∩ K,
then S = sH and T = sK = s(HK) = (sH)K = S K). Moreover S = sH ∈ T /H. Thus
G/H = ⊍ T /H
T ∈G/K
By (b) ∣T /H∣ = ∣K/H∣ and so
∣G/H∣ = ∑ ∣T /H∣ = ∑ ∣K/H∣ = ∣G/K∣ ⋅ ∣K/H∣
T ∈G/K
T ∈G/K
# 5. Let G be a group and a ∈ G. Recall that ia is the function defined by
ia ∶ G → G, g → ag
(a) Let Aut(G) be the set of automorphism of G. Show that Aut(G) is a subgroup of Sym(G).
(b) Show that ia is an automorphism of G.
(c) Show that the map Φ ∶ G → Aut(G), a → ia is a homomorphism.
(d) Let φ ∈ Aut(G). Show that iφ(a) = φ ○ ia ○ φ−1 .
(e) Let Inn(G) ∶= {ia ∣ a ∈ G}. Show that Inn(G) is a normal subgroup of Aut(G).
(f) Show that Inn(G) ≅ G/Z(G).
Proof. (a) idG is certainly an automorphism of G and so 1Sym(G) = idG ∈ Aut(G). Let α, β ∈ Aut(G). By
definition of an isomorphism, there exists a homomorphism γ ∶ G → G with α ○ γ = idG = γ ○ α. Thus
α−1 = γ ∈ Aut(G). Also for all a, b ∈ G:
(α ○ β)(ab) = α(β(ab)) = α(β(a)β(b)) = α(β(a))α(β(b)) = (α ○ β)(a)(α ○ β)(b)
and so α ○ β is a homomorphism. Since α ○ β is bijection, α ○ β ∈ Aut(G). Hence Aut(G) is a subgroup of
Sym(G).
(b) and (c): Let b, c ∈ G. Then
ia (bc) = abca−1 = aba−1 aca−1 = ia (b)ia (c)
and so ia is a homomorphism.
By Example 1.6.7, the map
Φ ∶ G → Sym(G), a → ia
is a (well defined) homomorphism. In particular, ia−1 is an inverse of ia and so ia is an automorphism. Hence
Φ(G) ⊆ Aut(G) and φ is also a homomorphism from G to Aut(G).
(d) Let b ∈ G. Since φ is a bijection, b = φ(c) for some c ∈ G. Then c = φ−1 (b). We compute
iφ(a) (b) = φ(a)bφ(a)−1 = φ(a)φ(c)φ(a−1 ) = φ(aca−1 ) = φ(ia (c)) = φ(ia (φ−1 (b))) = (φ ○ ia ○ φ−1 )(b)
and so iφ(a) = φ ○ ia ○ φ−1 .
(e) Note that Inn(G) = ImΦ and so Inn(G) is a subgroup of Aut(G). By (d)
φ
ia = iφ(a) ∈ Inn(G)
for all a ∈ G and φ ∈ Aut(G) and so Inn(G) ⊴ Aut(G).
(f) Note that Inn(G) = Gc , there c is the action of G on G by conjugation. By Example 1.6.11, G/Z(G) ≅
c
G and so (f) is proved.
# 6. Let E = (P, L, R) be a projective plane of order 2. Let A be a point and l a line incident with A. Put
S ∶= {α ∈ Aut(E) ∣ α(A) = A and α(l) = l}
(a) Show that ∣S ∣ = 8.
(b) Show that S is a subgroup of Aut(E).
(c) Show that ∣S ∣ is not abelian.
(d) Is S normal in Aut(E)?
Proof. Put G = Aut(E) and note that G acts on P, L and R.
(a) Let B be a point incident with l and distinct from A. Let C be point not incident with l. Let
N be the set of triples of non-collinear points. By Lemma 1.3.7 the function
G → N,
α → (α(A), α(B), α(C))
is a bijection. Let φ ∶ N → G be the inverse of this function.
1○ .
Let (Ã, B̃, C̃) ∈ N and α = φ(Ã, B̃, C̃). Then α ∈ S if and only if à = A, B̃ ∈ {B, A + B} and
C̃ ∈ P ∖ {A, B, A + B}.
Since α(A) = Ã, α(A) = A if and only if à = A. So we may assume that A = Ã. Then α(l) = α(AB) =
α(A)α(B) = A B̃. Thus α(l) = l if and only if B̃ is incident with l. Since B̃ ≠ Ã = A, this holds if and only if
B̃ ∈ {B, A + B}. C̃ just needs to be point not incident with à B̃ = l and so (1○ ) is proved.
Since there are two choices for B̃ and four choices for C̃ in (1○ ), we conclude that ∣S ∣ = 8.
(b) S = StabG ({A, l}) and so S is a subgroup of G by Lemma 1.6.10
(c) Let α = φ(A, B, B + C) and β = φ(A, A + B, C). By (1○ ) , α ∈ S and β ∈ S . We have
α(β(C)) = α(C) = B + C and β(α(C)) = β(B + C) = β(B) + β(C) = (A + B) + C
Since (A + B) + C ≠ B + C this shows α ○ β ≠ β ○ α and so S is not abelian.
(d) Put γ = φ(B, A, C) and α = φ(A, A + B, C). Then γ(A) = B and γ(l) = γ(AB) = γ(B)γ(A) = AB = l.
Thus by 1.6.9(f)
γ
S = γStabG ({A, l}) = StabG ({γ(A), γ(l)}) = StabG ({B, l})
Note that α(B) = A + B ≠ B and so α ∉ StabG ({B, l} = γS . By (1○ ), a ∈ S and so S ≠ γS . Thus S ⋬ G.
# 7. Let I be set, G ≤ Sym(I) and α ∈ Aut(G). Let ρ ∶ G → Sym(I), g → g be the inclusion map. Show that the
actions ∗ρ and ∗ρ○α of G on I are isomorphic if and only if there exists π ∈ Sym(I) with α(g) = πg for all
g ∈ G.
Proof. By definition ∗ρ and ∗ρ○α are isomorphic if and only there exists a G-equivariant bijection π from the
G-set (I, ∗ρ ) to the G-set (I, ∗ρ○α ). Let π ∶ I → I be a bijection, that is π ∈ Sym(I). Then
π is G-equivariant with respect to ∗ρ and ∗ρ○α
⇐⇒
π(g ∗ρ i) = g ∗ρ○α π(i)
for all g ∈ G, i ∈ I
⇐⇒ π(ρ(g)(i)) = ((ρ ○ α)(g))(π(i))
for all g ∈ G, i ∈ I
⇐⇒
π(g(i) = α(g)(π(i))
for all g ∈ G, i ∈ I
⇐⇒
π ○ g = α(g) ○ π
for all g ∈ G
⇐⇒
π ○ g ○ π−1 = α(g)
for all g ∈ G
⇐⇒
π
g = α(g)
for all g ∈ G
# 8. Let G be a group acting transitively on a set I and i ∈ I. Put Gi = StabG (i). Let H ≤ G. Show that H acts
transitively on I if and only if G = HGi .
Proof. Since G is transitive on I Lemma 1.6.15 show that for all j ∈ I there exists g ∈ G with j = gi. So
(∗)
I = {gi ∣ g ∈ G}
Thus
H acts transitively on G
⇐⇒
For all j ∈ I there exist h ∈ H with j = hi
− Lemma 1.6.15
⇐⇒
For all g ∈ G there exists h ∈ H with gi = hi
− (*)
−1
⇐⇒
For all g ∈ G there exists h ∈ H with h gi = i
⇐⇒
For all g ∈ G there exists h ∈ H with h−1 g ∈ Gi
⇐⇒
−1
For all g ∈ G there exist h ∈ H, k ∈ Gi with h g = k
⇐⇒
For all g ∈ there exist h ∈ H, k ∈ Gi with g = hk
⇐⇒
G = HGi
− Gi = StabG (i)
# 9. Let E = (P, L, R) be a projective plane of order 2 and G = Aut(E). Let l be a line, A the set of points incident
with l and B the set of points not incident with l. Observe that G acts on P and on L via αx = α(x) for α ∈ G
and x ∈ P ∪ L. Put H = StabG (l). Prove that
(a) G acts transitively on L.
(b) ∣H∣ = 24.
(c) H = NG (A).
(d) H acts faithfully on B.
(e) H ≅ Sym(B).
(f) Let β ∈ Sym(B). Then there exists α ∈ H with α∣B = β.
(g) Let γ ∈ Sym(A). Then there exists α ∈ H with α∣A = γ
(h) H/StabH (A) ≅ Sym(A).
(i) Let P be a point and put K = StabG (P). Then K ≅ H, but there does not exists g ∈ G with gH = K.
Proof. Let A, B, C be pairwise distinct points such that A and B are incident with l and C is not incident with
l.
(a) Let l˜ be a line and let Ã, B̃, C̃ be pairwise distinct points such that à and B̃ are incident with l˜ and
C̃ is not incident with l.˜ By Lemma 1.3.7 there exists α ∈ G with α(X) = X̃ for X = A, B and C. Then
α(l) = α(AB) = α(A)α(B) = Ã B̃ = l.˜ So G acts transitively on L.
(b) Since G act transitively on L we have Gl = L and Lemma 1.6.18(c) gives
∣G/H∣ = ∣G/∣StabG (l)∣ = ∣Gl∣ = ∣L∣
By Lagrange’s Theorem ∣G∣ = ∣G/H∣∣H∣ = ∣∣L∣∣H∣. Since ∣G∣ = 168 and ∣L∣ = 7 we get ∣H∣ =
∣G∣
∣L∣
=
168
7
= 24.
(c) Let α ∈ H and P ∈ P. Then P is incident to l if and only if α(P) is incident to α(l) = l. Thus α(A) = A
and H ≤ N(A).
Let α ∈ NG (A). Then α(A) and α(B) are incident with l and so α(l) = α(AB) = α(A)α(B) = l. Hence
α ∈ H and H = NG (A).
(d) By (b) , A is an H-invariant subset of P. Thus also B = P ∖A is H-invariant. Let α ∈ StabH (B). Since
∣B∣ = 4, B is non-collinear and so there exists a triple of non-collinear point in B. Since α fixes this triple of
non-collinear points we conclude that α = 1. Thus H acts faithfully on B.
(e) and (f) : For h ∈ H let h∗ be the restriction of h to B and put H ∗ = {h∗ ∣ h ∈ H}. Since H acts faithfully
in B, Lemma 1.6.9 shows that H ≅ H ∗ . Hence ∣H ∗ ∣ = ∣H∣ = 24 = 4! = ∣Sym(A)∣. Since H ∗ ≤ Sym(A), we
conclude that H ∗ = Sym(A). Thus (e) and (f) hold.
(g) Let γ ∈ Sym(A). By Lemma 1.3.7 there exists α ∈ G with α(A) = γ(A), α(B) = γ(B) and α(C) = C.
Thus α(l) = α(A)α(B) = γ(A)γ(B) = l, α ∈ H and both α(A + B) and γ(A + B) are the unique point in A
distinct from α(A) and α(B)). Thus γ is the restriction of α to A.
(h) For h ∈ H let h∗ be the restriction of h to A and put H ∗ = {h∗ ∣ h ∈ H}. By (g) H ∗ = Sym(A). By
Lemma 1.6.9 H/StabH (A) ≅ H ∗ and so (h) holds.
(i) Let B ∗ be the set of lines not incident with P. Let E ∗ = (L, P, R∗ ) be the dual plane of E and note that
G = Aut(E) = Aut(E ∗ ). Since P is a line in E ∗ , (e) applied to E ∗ in place of E, shows that K ≅ Sym(B ∗ ).
Since ∣B∣ = 4 = ∣B ∗ ∣ we conclude that K ≅ Sym(4) ≅ H and the first part of (i) holds. By (f) H does not fix
a point in B and by (g), H does not fix point on B. So H as no fixed-point. Let α ∈ G. By Lemma 1.6.9 (f),
α
P = StabG (α(P)). Since H does not fix α(P) we get H ≰ αP and H ≠ αP.