Test #3
Math 3110
Name:
April 10th , 2015
ANSWER KEY
Be sure to show your work!
1. (15 points) Getting things in order. . .
(a) Let G = D3 ⊕ Q where D3 = hx, y | x3 = 1, y 2 = 1, xyxy = 1i = {1, x, x2 , y, xy, x2 y} is a dihedral group and
Q = {±1, ±i, ±j, ±k} is the group of quaternions.
The order of G is |G| = |D3 | · |Q| = 6 · 8 = 48
Does G have an element of order 12? If so, give an example. If not, explain why not.
Yes, x has order 3 in D3 and i has order 4 in Q. Therefore, |(x, i)| = lcm(|x|, |i|) = lcm(3, 4) = 12
(b) Let G be a group with subgroups H and K such that H ⊆ K ⊆ G. In addition, suppose that |H| = 2 and |G| = 30.
What are the possible orders of K?
By Lagrange’s Theorem, |H| |K| and |K| |G|. This implies 2 |K| (i.e. |K| is even) and |K| 30. Because of this, the
possible orders of K are 2,6,10, or 30
2. (10 points) Let ϕ : G → H be a homomorphism between two groups G and H. Prove that ker(ϕ) / G.
[You may not assume that the kernel is a subgroup – prove this as well.]
K = Ker(ϕ) = {g ∈ G | ϕ(g) = e}
We need to run the normal subgroup test. . .
• ϕ(e) = e. This implies e ∈ K. Therefore, K is nonempty.
• Suppose a, b ∈ K. Then ϕ(a) = e and ϕ(b) = e. So ϕ(ab) = ϕ(a)ϕ(b) = e · e = e. Therefore, ab ∈ K.
• Suppose a ∈ K. Then ϕ(a) = e. This implies ϕ(a−1 ) = ϕ(a)−1 = e−1 = e, so a−1 ∈ K.
Therefore, K is a subgroup of G.
• Suppose g ∈ G and x ∈ K. Then ϕ(x) = e and so ϕ(gxg −1 ) = ϕ(g)ϕ(x)ϕ(g −1 ) = ϕ(g) · e · ϕ(g)−1 = e and so gxg −1 ∈ K
Thus, K is a normal subgroup of G.
3. (25 points) Quotients
(a) Given: H = {(1), (12)(34), (13)(24), (14)(23)} is a normal subgroup of S4 .
S4
The order of is
H
S4
The identity of is
H
4!
24
|S4 |
=
=
=6
|H|
4
4
H = (1)H
S4
is
The order of (1234)H in H
2
.
((1234)H)−1 =
.
.
(1234)−1 H = (1432)H
[= (1234)H]
The size of the set (1234)H is
4
.
.
Scratch work:
Notice that ((1234)H)2 = (1234)2 H = (12)(34)H = H since (12)(34) ∈ H, which implies |(1234)H| = 2 (order as an
element of the quotient group)
Also, all cosets have the same size, so since |H| = 4, (1234)H also has 4 elements [here |H| means the size of the set H,
not the order of H as an element of the quotient. In the quotient H is the identity so |H| = 1]
Now, ((1234)H)−1 = (1234)−1 H = (1432)H = {(1432)(1), (1432)(12)(34), (1432)(13)(24), (1432)(14)(23)}
= {(1432), (24), (1234), (13)}. Thus ((1234)H)−1 = (1234)H since both (1234) and (1432) belong to (1432)H. Another
way to see that ((1234)H)−1 = (1234)H is to notice that the order of (1234)H is 2, so it’s its own inverse!
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Z20
(b) Consider where H = h4i = {0, 4, 8, 12, 16}. List all of the cosets (and
H
Cayley table for this quotient group.
20
|Z20 |
=
= 4, so there should be 4 cosets.
Notice that [Z20 : H] =
|H|
5
Cayley Table:
• H = h4i = {0, 4, 8, 12, 16} (= 4 + H = 8 + H etc.)
H
• 1 + H = {1, 5, 9, 13, 17} (= 5 + H = 9 + H etc.)
H
H
1+H
1+H
• 2 + H = {2, 6, 10, 14, 18} (= 2 + H = 6 + H etc.)
2+H 2+H
• 3 + H = {3, 7, 11, 15, 19} (= 3 + H = 7 + H etc.)
3+H 3+H
their contents) of H in Z20 . Then make a
1+H
1+H
2+H
3+H
H
2+H
2+H
3+H
H
1+H
3+H
3+H
H
1+H
2+H
An example of computing in the Cayley table above: (2 + H) + (3 + H) = (2 + 3) + H = 5 + H = 1 + H.
∼ Z4 .
Also, notice (from the Cayley table) that Z20 /h4i =
4. (15 points) Let ϕ : Z6 → Z10 be defined by ϕ(x) = 5x.
(a) Show that ϕ is a homomorphism. [Do we need to prove that ϕ is well-defined?]
It is important to note that yes, we do need to prove that ϕ is well defined, since Z6 consists of equivalence classes and
our map, ϕ(x) = 5x, has been “defined” in terms of representatives of such equivalence classes.
Suppose x = y in Z6 . Then there is some k ∈ Z such that x = y+6k. This implies that 5x = 5y+30k, so 5x = 5y+10(3k).
Therefore, 5x = 5y in Z10 Thus ϕ is well defined.
Note that ϕ(x + y) = 5(x + y) = 5x + 5y = ϕ(x) + ϕ(y). Therefore, ϕ is a homomorphism.
(b) Compute the kernel and image of ϕ.
Let’s just compute ϕ(x) for x = 0, 1, . . . , 5:
0 7→ 0, 1 7→ 5, 2 7→ 0, 3 7→ 5, 4 7→ 0, 5 7→ 5.
So Ker(ϕ) = {x ∈ Z6 | ϕ(x) = 0} = {0, 2, 4} = h2i and ϕ(Z6 ) = {0, 5} = h5i.
(c) When applied to this ϕ, what does the first isomorphism theorem tell us?
Z6
Z6 ∼
∼
The first isomorphism theorem tells us that = ϕ(Z6 ), so = h5i (⊂ Z10 )
Ker(ϕ)
h2i
G
∼H
5. (10 points) Let H, K / G and G = HK. The second isomorphism theorem says that =
G
in class using the function ϕ : H → defined by ϕ(x) = xK.
K
K
H ∩K
. We proved this
(a) What needs to be proven about ϕ to establish this theorem?
We need to show that ϕ is an onto homomorphism. Then we need to show that Ker(ϕ) = H ∩ K. Then, by applying
H
H
G
∼
∼
the first isomorphism theorem, we get = ϕ(H), and so = Ker(ϕ)
H ∩K
K
Note: We don’t need to check if ϕ is well-defined since ϕ has elements of H as inputs (not representatives of equivalence
classes).
(b) Determine the kernel of ϕ.
Recall that K is the identity in G/K, so
Ker(ϕ) = {x ∈ H | ϕ(x) = K} = {x ∈ H | xK = K} = {x ∈ H | x ∈ K} = H ∩ K
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6. (25 points) Finite Abelian Groups
(a) List all of the non-isomorphic abelian groups of order 36 = 22 32 . Circle any that are cyclic.
The exponent “2” can be partitioned in two ways: 2 or 1 + 1. So there are two abelian groups of order 22 = 4 (up to
isomorphism): Z4 and Z2 ⊕ Z2 . Likewise, there are two abelian groups of order 32 = 9 (up to isomorphism): Z9 and
Z3 ⊕ Z3 . Therefore, we can build up four non-isomorphic abelian groups from these pairs.
Z4 ⊕ Z9 ∼
= Z36
Z2 ⊕ Z2 ⊕ Z9 ∼
= Z2 ⊕ Z18
Z4 ⊕ Z3 ⊕ Z3 ∼
= Z3 ⊕ Z12
Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ∼
= Z6 ⊕ Z6
[Not cyclic since the largest order appearing is lcm(2, 2, 9) = lcm(2, 18) = 18 6= 36.]
[Not cyclic since the largest order appearing is lcm(4, 3, 3) = lcm(3, 12) = 12 6= 36.]
[Not cyclic since the largest order appearing is lcm(2, 2, 3, 3) = lcm(6, 6) = 6 6= 36.]
(b) How many non-isomorphic abelian groups of order 25,508,082,600 are there?
Note: 25,508,082,600 = 23 34 52 71 113 132 and there are 5 non-isomorphic abelian groups of order 34 = 81. ,
Let p(k) be the number of partitions of k. Then we have p(1) = 1, p(2) = 2, p(3) = 3, and p(4) = 5. so the number of
non-isomorphic abelian groups of order 25 is p(3) · p(4) · p(2) · p(1) · p(3) · p(2) = 3 · 5 · 2 · 1 · 3 · 2 = 180 . I think I’ll skip
listing them. ,
(c) Are the groups Z6 ⊕ Z20 ⊕ Z9 and Z30 ⊕ Z36 isomorphic? Explain your answer.
Yes. Zn ⊕ Zm ∼
= Znm if and only if n and m are relatively prime. Since we can pull apart and put back together the
various moduli to get from one group to the other, they are isomorphic.
We see that Z6 ⊕ Z20 ⊕ Z9 ∼
= Z2 ⊕ Z3 ⊕ Z4 ⊕ Z5 ⊕ Z9 ∼
= Z2 ⊕ Z3 ⊕ Z5 ⊕ Z4 ⊕ Z9 ∼
= Z30 ⊕ Z36
(d) Is the group Z2 ⊕ Z21 ⊕ Z15 cyclic? Explain you answer.
No, it is not cyclic. If it were cyclic, we should be able to rewrite it as Zn for some n (actually n = 630). But we can’t
since 21 and 15 aren’t relatively prime.
Note that Z2 ⊕ Z21 ⊕ Z15 ∼
= Z2 ⊕ Z3 ⊕ Z7 ⊕ Z3 ⊕ Z5 ∼
= Z3 ⊕ Z210 (the largest element order is lcm(2, 21, 15) = lcm(3, 210) =
210 6= 630, so not cyclic).
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