MT310 Homework 9
Solutions
Due Saturday, April 17 by midnight
Exercise 1. Suppose G is a nonabelian group of order p3 . We have seen that the center Z = Z(G)
has order p. Let a ∈ G, but a ∈
/ Z.
a) Prove that for all b ∈ G, we have a−1 bab−1 ∈ Z. (See exam 2, problem 3).
b) Prove that the map fa : G → Z given by fa (b) = a−1 bab−1 is a surjective group homomorphism,
with ker fa equal to the centralizer CG (a) of a.
c) Prove that the conjugacy class of a in G is the coset aZ.
d) How many conjugacy-classes of each order does G have?
Proof. a) In exam 2, you showed that if G/H is abelian then xyx−1 y −1 ∈ H for all x, y ∈ G. In this
case, we have G/Z ' Zp × Zp abelian, so a) holds.
b) We compute
fa (bc) = a−1 (bc)a(bc)−1 = a−1 bcac−1 b−1 = a−1 ba · a−1 cac−1 · b−1 = a−1 ba · fa (c) · b−1 .
Since fa (c) ∈ Z, by part a), we have
fa (bc) = a−1 bab−1 · fa (c) = fa (b)fa (c),
so fa is a group homomorphism. Since a ∈
/ Z, the image of fa is a nontrivial subgroup of Z. Since
|Z| = p, the image of fa equals Z. That is, fa is surjective. Its kernel is determined as follows.
ker fa = {b ∈ G : a−1 bab−1 = e} = {b ∈ G : bab−1 = a} = CG (a).
c) The conjugacy-class of a is {bab−1 : b ∈ G} = {a · fa (b) : b ∈ G} = a{fa (b) : b ∈ G} = aZ.
d) From c), the p3 − p noncentral elements of G are partitioned into conjugacy-classes of size p.
And each element of Z is alone in its conjugacy class. Hence there are p2 − 1 conjugacy classes of size
p and p conjugacy classes consisting of one element.
Exercise 2. In homework 6, you showed that the set of matrices
1 0
1 1
0 1
0 0
F =
,
,
,
0 1
1 0
1 1
0 0
under matrix addition and multiplication, is a field. We have also seen that Z2 [x] has a unique quadratic
polynomial, namely x2 + x + 1. Prove that
Z2 [x]/(x2 + x + 1) ' F.
1 1
Proof. The matrix α =
satisfies α2 + α + 1 = 0. The polynomial x2 + x + 1 is irreducible in
1 0
Z2 [x], so it generates the kernel of the homomorphism
φα : Z2 [x] → F,
φα (f ) = f (α).
Since F = {0, 1, α, α2 }, the ring homomorphism φα is surjective. From the First Isomorphism Theorem,
we get an isomorphism
∼
φ̄α : Z2 [x]/(x2 + x + 1) −→ F.
Exercise 3. Let F be a field, and let f (x) ∈ F [x] be a polynomial of degree n ≥ 1, generating the
ideal (f ) = {gf : g ∈ F [x]}. Prove that each coset in F [x]/(f ) contains a unique polynomial of degree
< n.
Proof. The ideal (f ) is unchanged if we divide f by its coefficient of xn . Hence we may assume that f
is monic:
f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 .
It suffices to prove by induction on k ≥ 0 that
xn+k ∈ hk (x) + (f ),
where hk (x) ∈ F [x] has degree < n. For k = 0, we have
xn = −an−1 xn−1 − · · · − a1 x − a0 + f (x),
so the claim holds, with h0 (x) = −an−1 xn−1 − · · · − a1 x − a0 . Assume that xn+k−1 ∈ hk−1 (x) + (f ),
where hk−1 (x) ∈ F [x] has degree < n. Write
hk−1 (x) = cn−1 xn−1 + `(x),
where deg `(x) < n − 1. Then
xn+k ∈ xhk−1 (x) + (f ) = cn−1 xn + x`(x) + (f ) = cn−1 h0 (x) + x`(x) + (f ).
Thus, we have xn+k ∈ hk (x) + (f ), where hk (x) = cn−1 h0 (x) + x`(x). Since h0 (x) and x`(x) both have
degree < n, it follows that deg hk < n, so the claim is proved by induction.
As for uniqueness, suppose g+(f ) = h+(f ) with both g and h having degrees < n. Then g−h ∈ (f ),
so f divides the polynomial g − h of degree < n. Since deg f = n, this can only hold if g − h = 0, that
is, if g = h.
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