MATH 251 Homework 2 Solutions 1. Prove that no two of the ...

MATH 251 Homework 2 Solutions
1. Prove that no two of the multiplicative groups (Q ∗ , ·), (R ∗ , ·) and (C ∗ , ·) are isomorphic.
Proof : Since Q ∗ is countable and R ∗ and C ∗ are uncountable and an isomorphism is a bijection, we see that
(R ∗ , ·) ≁ = (Q ∗ , ·) ≁ = (C ∗ , ·).
Suppose that θ : C ∗ → R ∗ is an isomorphism. Note that i is an element in C of order four. Then x := i θ is
an element in R of order four. So x 4 = 1. Hence x = 1 or x = −1. In particular, the order of x is not four, a
contradiction. So (C ∗ , ·) ≁ = (R ∗ , ·).
✷
2. This exercise is about S 4 .
(a) Write down how many elements S 4 has of each cycle type (like elements of the form (ab) or (abc) etc.).
(b) Prove that Z(S 4 ) = {e} (hint : for each cycle type σ , find an element in S 4 that does not commute with
σ).
Solution : (a) We easily get the following :
e : 1
(ab) :
( 4
6 =
2)
( ) 4 3!
(abc) : 8 =
3 3
(abcd) : 6 = 4!
4
(ab)(cd) : 3 = 1 ( 4
2 2)
(b) Note that
(ab)(abc) = (ac) and (abc)(ab) = (bc)
So (ab), (abc) /∈ Z(S 4 ). We also have that
(abcd)(ab)(cd) = (bd) and (ab)(cd)(abcd) = (ac)
So (abcd), (ab)(cd) /∈ Z(S 4 ).
Hence Z(S 4 ) = {e}.
✷
3. Let K be a field. Prove that SL(n, K) ✂ GL(n, K) and that GL(n, K)/SL(n, K) ∼ = K ∗ .
Proof :
Consider the map
θ : GL(n, K) → K ∗ : A → det(A)
Note that det(A) ≠ 0 and so det(A) ∈ K ∗ if A ∈ GL(n, K).
• θ is a homomorphism
Let A, B ∈ GL(n, K). Using a well-known property of determinants, we get that
θ(AB) = det(AB) = det(A) det(B) = θ(A)θ(B)
1

• θ is onto
Pick a ∈ K ∗ . Put
Then A ∈ GL(n, K) and θ(A) = det(A) = a.
• Ker θ = SL(n, K)
Let A ∈ GL(n, K). Then
Hence Ker θ = SL(n, K).
⎡
⎤
a 0 0 · · · 0
0 1 0 · · · 0
A =
0 0 1 · · · 0
⎢
⎣
.
. . . ..
⎥
0 ⎦
0 0 0 · · · 1
A ∈ Ker θ ⇐⇒ θ(A) = 1 ⇐⇒ det(A) = 1 ⇐⇒ A ∈ SL(n, K)
It follows that SL(n, K) = Ker θ ✂ GL(n, K). By the First Isomorphism Theorem, we get
GL(n, K)/SL(n, K) = GL(n, K)/Ker θ ∼ = Im θ = K ∗
4. Let G be a group, H ≤ G and N ✂ G. Prove that HN = NH ≤ G.
Proof : Pick y ∈ HN. Then y = hn for some h ∈ H and some n ∈ N. Since N ✂ G, we have that hN = Nh.
So y = hn ∈ hN = Nh ⊆ NH. Hence HN ⊆ NH.
Pick y ∈ NH. Then y = nh for some h ∈ H and some n ∈ N. Hence y = nh ∈ Nh = hN ⊆ HN. Hence
NH ⊆ HN.
So HN = NH.
Pick x 1 , x 2 ∈ HN. Then x 1 = h 1 n 1 and x 2 = h 2 n 2 for some h 1 , h 2 ∈ H and some n 1 , n 2 ∈ N. Hence
So HN ≤ G by the Subgroup Test.
x 1 x −1
2 = (h 1 n 1 )(h 2 n 2 ) −1 = h 1 n 1 n −1
2 h−1 2 ∈ HNH = HHN = HN
✷
5. Let G be a group and H ≤ G such that [G : H] = 2. Prove that H ✂ G.
Proof : Pick g ∈ G. If g ∈ H, then clearly gH = H = Hg. So assume that g /∈ H. Then {H, gH} (resp.
{H, Hg}) are the two left (resp. right) cosets of H in G. Since the cosets form a partition of G, we have that
gH = G \ H = Hg. In either case, we get that gH = Hg. So H ✂ G.
✷
6. Let G be a group. Prove that Z(G) char
✂ G.
Proof :
Let θ ∈ Aut(G). It is enough to show that θ(Z(G)) ⊆ Z(G) by Proposition 1.19b.
Pick x ∈ Z(G). Let g ∈ G. Since θ : G → G is an isomorphism (and hence onto), we have that g = θ(h) for
some h ∈ G. Using the fact that x ∈ Z(G) and θ is a homomorphism, we find
θ(x)g = θ(x)θ(h) = θ(xh) = θ(hx) = θ(h)θ(x) = gθ(x)
Since g was arbitrary, we get that θ(x) ∈ Z(G). So θ(Z(G)) ⊆ Z(G).
✷
7. True/False : Let G be a group and A ✂ B ✂ G. Then A ✂ G.
Solution : False. Let G = D 4 , B = {e, r 2 , s, r 2 s} and A = {e, s}. Then A, B ≤ G. Since [G : B] = [B : A] = 2,
we get that A ✂ B ✂ G. But r −1 sr = sr 2 /∈ A. Hence A is not a normal subgroup of G.
✷
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