Download HW 1

Math 210A: Algebra, Homework 1
Ian Coley
October 9, 2013
Problem 1. Let a1 , a2 , . . . , an be elements of a group G. Define the product of the ai ’s by
induction: a1 a2 · · · an = (a1 a2 · · · an−1 )an .
(a) Prove that
a1 a2 · · · an b1 b2 · · · bm = (a1 a2 · · · an )(b1 b2 · · · bm ).
(b) Prove that a1 a2 · · · an is equal to the product of the ai ’s with the parentheses inserted
arbitrarily.
Solution.
(a) We proceed by induction on m. Clearly a1 · · · an b1 = (a1 · · · an )(b1 ). Now assume the
case of m = k. Then
a1 · · · an b1 · · · bk bk+1 = (a1 · · · an b1 · · · bk )bk+1 = (a1 · · · an )(b1 · · · bk )bk+1
= (a1 · · · an )(b1 · · · bk bk+1 ).
This completes the argument.
(b) Write a1 a2 · an = a1 · · · am · am+1 · · · an for any 1 < m < n. Then by 1(a), a! · · · an =
(a1 · · · am )(am+1 · · · an ). We may continue this process for either of the terms, hence
the general product is equal to one with arbitrarily many parentheses inserted.
Problem 2.
(a) Prove that for any n ∈ N, the set of all complex nth roots of unity forms a group with
respect to complex multiplication. Show that this group is cyclic.
(b) Prove that if G is a cyclic group of order n and k divides n, then G has exactly one
subgroup of order k.
Solution.
(a) Let µ denote the nth roots of unity, and let ζ1 , ζ2 ∈ µ. Then (ζ1 ζ2 )n = ζ1n ζ2n = 1 ∈ C,
so ζ1 ζ2 ∈ µ as well. Further, 1n = 1 ∈ C so 1 ∈ µ. Thus µ is a subgroup under
multiplication in C× . Suppose µ is a primitive nth root unity, i.e. ord(ζ) = n. Then
|hζi| = n = |µ| and hζi ⊂ µ, so hζi = µ so µ is cyclic.
1
(b) First, we claim that a subgroup of a cyclic group is still cyclic. Let H ⊂ G = hai be
a subgroup. Let m be the smallest positive integer such that am ∈ H. We claim that
ham i = H. Let h ∈ H, and let h = as . By the division algorithm, we may write
s = mq + r
where 0 ≤ r < m. Therefore h = (am )q ar . Since am ∈ H, (am )−q ∈ H, and since H is
closed we have (am )−q · h = ar ∈ H. Since r < m by assumption, this would contradict
minimality if r > 0, so we have r = 0. We conclude h = (am )q for all h ∈ H, hence am
generates H.
0
Now let ham i and ham i be two subgroups of order k. Without loss of generality, we
0
may let m = n/k, since this subgroup does have k elements. Since |ham i| = d, we
0
know (am )d = e, so n | m0 d. Therefore write
qn = m0 k =⇒ m0 = q ·
n
= qm.
k
0
0
Therefore ham i ⊂ ham i and since these subgroups have equal order, ham i = ham i.
Hence there is only one subgroup for each k | n.
Problem 3.
(a) Show that if K and N are two finite subgroups of G of relatively prime orders, then
K ∩ N = {e}.
(b) Show that if a group G has only finitely many subgroups, then G is finite.
Solution.
(a) Let |K| = m and |N | = n so that (m, n) = 1. Then since K ∩ N is a subgroup of
both K and N , we have |K ∩ N | divides both m and n (again by Lagrange), hence it
divides their gcd. But (m, n) = 1, so |K ∩ N | = 1, so K ∩ N = {e}.
S
(b) Note that G = hgi, ranging over all cyclic subgroups of G. This is because g ∈ hgi for
every g ∈ G. Suppose every element of g has finite order. Then hgi is finite for every
g ∈ G. Since there are only a finite number of subgroups, G is a finite union of finite
subgroups, hence is finite itself. Suppose now that g ∈ G is of infinite order. Then
hgi ∼
= Z. Z has infinitely many subgroups (namely of the form nZ), and so hgi has
infinitely many subgroups as well, which contradicts our assumption. This completes
the proof.
Problem 4.
(a) Let G be a group of order n. Show that an = e for all a ∈ G.
(b) Prove that a group G is cyclic if and only if there is an element a in G with ord(a) = |G|.
(c) Show that every group of prime order is cyclic.
2
Solution.
(a) Assume a ∈ G such that an 6= e. Because |G| = n, we cannot have ord(a) > n, else
{e, a, . . . , an } would all be distinct and this contradicts the order of G. Let ord(a) =
m ≤ n. Then |hai| = m, the cyclic subgroup generated by a. By Lagrange’s theorem,
|hai| | |G| =⇒ m | n. Therefore write n = dm. Then
an = (am )d = ed = e.
This completes the proof.
(b) If G is cyclic, then G = hai. Therefore G = {1, a, . . . , an−1 }, where each power of a is
different. Further, an = 1 by the previous statement. Hence ord(a) = n. Conversely,
suppose there exists a with order n. Then |hai| = |G| and hai ⊂ G, so hai = G and
hence G is cyclic.
(c) Suppose |G| = p with p a prime. Let a 6= e. Then hai is a subgroup of G, so by
Lagrange’s theorem |hai| divides p. Since p is prime, |hai| = 1 or p, and it is not 1 by
assumption that a 6= e. Since ord(a) = |G|, by the previous statement G is cyclic.
Problem 5.
(a) Show that if a2 = e for all elements a of a group G, then G is abelian.
(b) Prove that if G is a finite group of even order, then G contains an element a 6= e such
that a2 = e.
(c) Show that every subgroup of index 2 is normal.
Solution.
(a) If a2 = e for all a ∈ G, then let a, b ∈ G be two elements. Then
aba−1 b−1 = abab = (ab)(ab) = e ⇐⇒ ab = ba,
where the implication follows by left multiplication by b and then a. Therefore G is
abelian.
(b) We may associate for every a ∈ G its inverse a−1 ∈ G. If a2 6= e for all a, then
a 6= a−1 , hence this pairing exhausts all non-identity elements. Therefore the number
of elements of the group is 2n + 1, where n is the number of pairings and 1 is the
identity element e. However, we assumed G was a group of even order, so this is a
contradiction. Hence G contains an element of order 2.
(c) Let H ⊂ G such that |G : H| = 2. Then there are two left cosets H and gH for any
g ∈ G \ H, and two right cosets H and Hg for any g ∈ G \ H. Since H 6= Hg, we must
have gH = Hg for all g ∈ G \ H, and this holds for g ∈ H as well. gH = Hg ⇐⇒
gHg −1 = H, so H is normal in G.
3
Problem 6. Find all groups (up to isomorphism) of order ≤ 5. What is the smallest order
of a non-cyclic group?
Solution. Up to isomorphism, there is only one trivial group. By 4(b), every group of prime
order is cyclic. By the fundamental theorem of cyclic groups, every cyclic group is unique
up to isomorphism. Therefore there is only one group each of order 2, 3, and 5.
However, for order 4, there is the cyclic group Z/4Z and the non-cyclic group V =
{1, a, b, ab} such that a2 = b2 = 1 and ab = ba. Suppose G is a different group of order 4 (up
to isomorphism). G is not cyclic by assumption, and by 5(b) G contains an element of order
2, say x. Since G does not contain any elements of order 4 and only the identity has order
1, it must contain two other elements of order 2, call them y, z. Then we must have xy = z
and xz = y. Therefore there exists an isomorphism ϕ : G → V such that x 7→ a, y 7→ b, and
z 7→ ab so G ∼
= V . Therefore there are only these two groups of order 4.
The smallest order of a non-cyclic group is because there is only one trivial group and
every group of prime order is cyclic by 4(b).
Problem 7. Find a non-normal subgroup in the symmetric group S3 .
Solution. Consider the subgroup H = {(1), (1 2)}. Note that (1 3)(1 2)(3 1) = (3 2) 6∈ H,
so H 6= (1 3)H(3 1). Therefore H is not normal.
Problem 8. Let n ∈ N. Show that the map
f : Q/Z → Q/Z,
f (a + Z) = na + Z
is a well-defined homomorphism. Find ker f and im f .
Solution. Let a + Z and b + Z be in Q/Z. Then
f ((a + Z) + (b + Z)) = f (a + b + Z) = n(a + b) + Z = na + nb + Z
= (na + Z) + (nb + Z) = f (a + Z) + f (b + Z).
Further, f (0 + Z) = n0 + Z = 0 + Z, so f is properly a homomorphism. Now let a + Z, a0 + Z
be two representatives of the same coset, so that a0 + b = a for b ∈ Z. Then
f (a + Z) = na + Z = n(a0 + b) + Z = na0 + nb + Z = na0 + Z = f (a0 + Z),
where nb + Z = Z because nb ∈ Z. Thus f is a well-defined homomorphism.
Now, if f (a+Z) = 0+Z, then we must have na ∈ Z. Thus ker f is all cosets representable
by a = p/n where p ∈ Z. For the image, fix the coset b ∈ Z. Then clearly f (b/n+Z) = b+Z,
and since b/n ∈ Q, we see every element of the codomain has a nontrivial preimage. Hence
im f = Q/Z.
Problem 9.
(a) Show that Z/6Z ∼
= Z/2Z × Z/3Z.
(b) Prove that nZ/mnZ ∼
= Z/mZ.
4
Solution.
(a) Define an map ϕ : Z/6Z → Z/2Z × Z/3Z by 1 7→ (1, 1). Since 1 is a generator for
Z/6Z, this defines the entire map. We claim this is an isomorphism. First, ϕ(0) =
ϕ(6 · 1) = 6 · (1, 1) = (0, 0). Second,
ϕ(a + b) = ϕ((a + b) · 1) = (a + b) · ϕ(1) = a · ϕ(1) + b · ϕ(1) = (a + b) · ϕ(1),
so ϕ is a homomorphism. Note that (1, 1) is a generator for Z/2Z × Z/3Z. Therefore
ϕ(a · 1) = a · ϕ(1) 6= b · ϕ(1) = ϕ(b · 1) for any a 6= b with 1 ≤ a, b ≤ 6. This shows that
ϕ is injective. Since |Z/6Z| = 6 = |Z/2Z × Z/3Z|, ϕ is surjective as well. Hence ϕ is
an isomorphism.
(b) Note that both groups are cyclic, viz. Z/mZ = h1i and nZ/mnZ = hni. Further,
they each have order m. We claim that ϕ : nZ/mnZ → Z/mZ given by n 7→ 1 is an
isomorphism. Since ϕ maps a generator to a generator, ϕ is injective and surjective.
Therefore ϕ is an isomorphism and we are done.
Problem 10. Let f : G → H be a surjective group homomorphism, let H 0 be a normal
subgroup in H. Show that G0 = f −1 (H 0 ) is a normal subgroup in G and G/G0 ∼
= H/H 0 .
Solution. We would like to show that xgx−1 ∈ G0 for every g ∈ G0 and g ∈ G. We know
that f (xgx−1 ) = f (x)f (g)f (x)−1 . Since f (g) ∈ H 0 and f (x), f (x)−1 ∈ H, we know that
f (x)f (g)f (x)−1 ∈ H 0 . Therefore xgx−1 ∈ f −1 (H 0 ) = G0 , so G0 is normal.
Now we claim f induces an isomorphism f 0 : G/G0 → H/H 0 . f 0 is a homomorphism
because f is. Since f −1 (H 0 ) = G0 , and H 0 is the identity element in H/H 0 , f 0 is injective.
Further, since f was surjective to begin with, the induced map f 0 is also surjective. Therefore
f is an isomorphism and we are done.
5