Math 546
Homework 3
Due Wednesday, February 8.
Contents of this handout:
1. 546 Problems. All students (students enrolled in 546 and 701I) are required to turn these in.
2. 701I Problems. Only students enrolled in 701I are required to turn these in. Students not
enrolled in 701I are welcome to turn these in as well for additional credit. I especially
welcome students looking for a challenge to attempt these.
3. Bonus Problems. All students (students in 546 and 701I) may turn these in for additional
credit. Some of these may be challenging.
4. Examples. I wrote solutions to certain non-assigned problems from the text. These may
be helpful for you when you are working on the problems and when you are writing your
solutions.
1
546 Problems.
4.16. Suppose that G = hxi is a cyclic group. Prove that G = hx−1 i.
4.21. Let G be a group, and let x, y ∈ G. Prove that o(xy) = o(yx).
Note: Let n be a positive integer. It suffices to show that (xy)n = e if and only if (yx)n = e.
(Make sure that you understand why this is sufficient.) See my solution to 4.20 below for a
similar argument.
4.22. Let G be an abelian group, and let x, y ∈ G. Suppose that x and y have finite order.
Show that
(a) xy has finite order and
(b) o(xy) | o(x)o(y).
Note: Suppose, in addition to the hypothesis of the problem, that gcd(o(x), o(y)) = 1. Then
we may conclude that o(xy) = o(x)o(y). This is the content of 4.23, which is an assigned
701I problem. (See below.)
However, you should be aware that o(xy) 6= o(x)o(y) in general. In particular, we can have
o(xy) 6= o(x)o(y) when G is non-abelian or when gcd(o(x), o(y)) 6= 1.
Note: You may find it helpful to see my solutions to Problems 5.1 c) and f) and 5.22 on how
to apply the subgroup criteria from Theorem 5.1 to determine whether or not a subset of a
group is a subgroup.
5.1. Each problem below presents a group G and a subset H ⊆ G. Determine whether or
not H is a subgroup of G.
d) G = (Q× , ·); H = Q+ = {q ∈ Q : q > 0}. (Recall that Q× = Q \ {0}.)
e) G = (Z8 , ⊕); H = {0, 2, 4}.
5.3. Define the set
a b
SL(2, R) := M =
: det M = ad − bc = 1 ⊂ GL(2, R).
c d
Prove that SL(2, R) is a subgroup of GL(2, R). It is called the special linear group of
degree 2 over R.
5.11. Let G be an abelian group, let n be a positive integer, and let
H := {x ∈ G : xn = e} ⊆ G.
Show that H is a subgroup of G.
2
701I Problems.
4.23. Let G be an abelian group, and let x, y ∈ G have finite order with gcd(o(x), o(y)) = 1.
Prove that o(xy) = o(x)o(y).
Note: You may use the result of 4.22 above that xy has finite order and that o(xy) | o(x)o(y).
Therefore, to prove the statement in 4.23, it suffices to show that o(x)o(y) | o(xy). It is likely
that you will need certain properties of the gcd to do this.
5.23. Let G be a group, and let g ∈ G. Define the centralizer of g in G, Z(g), to be
Z(g) = {x ∈ G : xg = gx}.
Hence, the centralizer of g in G is the set of elements in G which commute with g. Prove
that Z(g) is a subgroup of G.
3
Bonus Problems.
4.24. Let G be a group, and let x, y ∈ G. Suppose that x 6= e, that o(y) = 2, and that
yxy −1 = x2 . Find o(x).
4.25. Suppose that |G| is even. Show that there exists x 6= e in G with x2 = e.
5.25. Let G be a group, let a ∈ G, and let H ⊆ G be a subgroup. Define
aHa−1 = {aha−1 : h ∈ H}.
Prove that aHa−1 is a subgroup of G. It is the conjugate subgroup of H by a.
4
Examples.
4.20. Let G be a group, and let a ∈ G. Then b ∈ G is a conjugate of a if and only if there
exists x ∈ G with b = xax−1 . Let c ∈ G be any conjugate of a. Show that o(c) = o(a).
Proof: Since c is a conjugate of a, there exists y ∈ G with c = yay −1 . Therefore, we will
show that o(a) = o(yay −1 ). Let n be a positive integer. By the definition of order in a group,
it suffices to show that (yay −1 )n = e if and only if an = e. We reason as follows:
(yay −1 )n = e ⇐⇒
⇐⇒
⇐⇒
⇐⇒
(yay −1 )(yay −1 ) · · · (yay −1 ) = e
ya(y −1 y)a(y −1 y)a · · · a(y −1 y)ay −1 = e
yan y −1 = e
an = y −1 ey = e.
We note that the first equivalence follows from multiplying yay −1 by itself n times. .
5.1. Each problem below presents a group G and a subset H ⊆ G. Determine whether or
not H is a subgroup of G.
c) G = (Z, +); H = Z+ = {n ∈ Z : n > 0}.
Claim: Z+ is not a subgroup of Z under addition.
Proof: It suffices to show that one of the two subgroup criteria (closure, inverses) fails.
We claim that Z+ fails to have additive inverses. For example, 1 ∈ Z+ ⊂ Z has additive
inverse −1 ∈ Z, but −1 6∈ Z+ . Since Z+ fails to have additive inverses, Z+ is not a
subgroup of Z under addition. .
f) G = {(a, b) : a, b ∈ R} is a group under the binary operation of addition: for all (a, b),
(c, d) ∈ G, we have (a, b) + (c, d) = (a + c, b + d) ∈ G.
Let H = {(a, −a) : a ∈ R} ⊂ G.
Claim: H is a subgroup of G.
Proof: We first note that (0, 0) ∈ H, so H is non-empty.
It suffices to verify that H meets the subgroup criteria from Theorem 5.1:
(i) H is closed under addition: To see this, we observe, for all (a, −a), (b, −b) ∈ H
that (a, −a) + (b, −b) = (a + b, −a − b) = (a + b, −(a + b)) ∈ H.
(ii) H has additive inverses. To see this, let (a, −a) ∈ H, and note that (0, 0) ∈ H is
the identity in G. Then (−a, a) ∈ H and
(a, −a)+(−a, a) = (a+(−a), −a+a) = (0, 0) = (−a+a, a+(−a)) = (−a, a)+(a, −a).
Hence, (−a, a) ∈ H is the inverse of (a, −a), so H has inverses.
We conclude that H is a subgroup of G under addition. 5.22. Let G be a group. The center of G is
Z(G) = {z ∈ G : for all x ∈ G, we have zx = xz}
= {elements in G which commute with everything in G}.
Prove that Z(G) is a subgroup of G.
Proof: We first note that e ∈ Z(G), so we have Z(G) 6= ∅. We now verify that Z(G) meets
the subgroup criteria.
(i) Z(G) is closed. To prove this, we let a, b ∈ Z(G). Then for all x ∈ G, we have
ax = xa and bx = xb. We will show, for all y ∈ G, that (ab)y = y(ab). We compute
(ab)y = a(by) = a(yb) = (ay)b = (ya)b = y(ab).
We conclude that ab ∈ Z(G), so Z(G) is closed. We note that the second equality
follows since b ∈ Z(G), and the fourth equality follows since a ∈ Z(G).
(ii) Z(G) has inverses. To see this, we suppose that a ∈ Z(G). Then for all x ∈ G, we
have ax = xa. We note that a ∈ G; since G is a group, we have a−1 ∈ G. We will
show that a−1 ∈ Z(G): Let y ∈ G. We compute
ya−1 = (ay −1 )−1 = (y −1 a)−1 = a−1 y.
Since the arbitrary element y commutes with a−1 , we have a−1 ∈ Z(G), so Z(G) has
inverses. We note that the second equality follows since a ∈ Z(G).
We conclude that Z(G) is a subgroup of G.