Name:
MA407 Test 2
Miss Bosko
Show all work. Good luck!
(1) Prove or disprove that U (15) ∼
= Z8 . (8pts)
U (15) = {1, 2, 4, 7, 8, 11, 13, 14}. So both groups have the same
order. Also, both are abelian. Next, look at the order of each
of the elements. In U (15),
o(1) = 1, o(2) = 4, o(4) = 2, o(7) = 4, o(8) = 4, o(11) =
1, o(13) = 4, o(14) = 2. Since U (15) has no elements of order
8, it is not cyclic and cannot be isomorphic to Z8 .
(2) Let G = h5i and H = h20i be subgroups of Z.
(a) Why is H a normal subgroup of G? (2pts)
H is normal since G is abelian.
(b) Determine the factor group G/H. As in class, you will
need to list the cosets as well as the elements in the cosets.
(10pts)
G = {0, ±5, ±10, ±15, . . .} and H = {0, ±20, ±40, ±60, . . .},
so
G/H = {H, 5 + H, 10 + H, 15 + H} where
5 + H = {. . . , −55, −35, −15, 5, 25, 45, 65, . . .}
10 + H = {. . . , −50, −30, −10, 10, 30, 50, 70, . . .}
15 + H = {. . . , −45, −25, −5, 15, 35, 55, 75, . . .}
(c) Determine the orders of each of the elements in G/H (5pts)
o(H) = 0
o(5 + H) = 4 since (5 + H)4 = 20 + H = H
o(10 + H) = 2 since (10 + H)2 = 20 + H = H
o(15 + H) = 4 since (15 + H)4 = 60 + H = H
(d) G/H is isomorphic to what group? (You do not need to
prove this). (2pts)
Since H is finite, cyclic, and of order 4 it must be isomorphic to Z4
(3) Let G = D4 .
(a) Let H = {I, R2 }. List the left cosets of H in G. (4pts)
H = {I, R2 }, RH = {R, R3 }, V H = {V, V R2 }, (V R)H =
{V R, V R3 }
(b) Is H normal in G? Why or why not? (3pts)
H is normal in G since the left and right cosets are equal.
1
2
Specifically,
H = H, RH = HR, V H = HV, (V R)H = H(V R)
(4) Let G = S4 by the set of permutations on the set {1, 2, 3, 4}.
(a) Determine the elements of H = h(1234)i. Use Lagrange’s
Theorem to determine the number of distinct right cosets
of H in G. (5pts)
Notice that H = {(1234), (13)(24), (1432), (1)} and o(S4 ) =
|G|
4! = 24, the number of distinct right (or left) cosets |H|
=
24
= 6.
4
(b) Let ρ ∈ S4 . Determine the possibilities for ρ24 . (2pts)
By a corollary to Lagrange’s Theorem, ρ|G| = e. Thus,
ρ24 = (1).
(c) Suppose K ≤ G and J ≤ G where | K |= 3 and | J |= 4.
Determine the possibilities for the order of K ∩ J (3pts)
We know, K ∩ J is a subgroup of both K and J. By
Lagrange’s Theorem, | K ∩ J | divides the order of K and
the order of J. Thus, | K ∩ J |= 1.
(5) Let G = Z24 and H = h6i.
(a) Determine the elements in the factor group G/H. (2pts)
G/H = {H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H} where
H = {6, 12, 18, 0}, 1+H = {7, 13, 19, 1}, 2+H = {8, 14, 20, 2}, 3+
H = {3, 15, 21, 3}, 4 + H = {10, 16, 22, 4}, 5 + H =
{11, 17, 23, 5}
(b) What is the order of the element 15 + h6i in the factor
group G/H? Be sure to show work supporting your answer.
(4pts)
(15 + h6i)2 = (15 + h6i) + (15 + h6i) = 30 + h6i = 6 + h6i =
h6i = e. Thus, o(15 + h6i) = 2
3
Name:
MA407 Test 2
Miss Bosko
Show all work. You may use your notes, textbook, homework assignments, or worksheets from this class. No other resources are permitted.
(1) Prove that R+ is isomorphic to R by finding a mapping from
R+ to R and then proving that it is an isomorphism. It may be
helpful to draw a picture on the Cartesian plane. (5pts)
An example of a function with domain R+ and range R is φ
defined by φ(x) = ln(x). Now, we prove φ is an isomorphism.
To show 1 − 1, suppose ln(x) = ln(y). Then eln(x) = eln(y) and
x = y.
To show onto, let y ∈ R. Then φ(ey ) = ln(ey ) = y and ey ∈ R+ .
To show homomorphism, φ(xy) = ln(xy) = ln(x) + ln(y) =
φ(x) + φ(y).
Thus, φ is an isomorphism.
(2) Let H ≤ G and let a, b, ∈ G. Prove or give a counterexample
for each of the following statements. (5pts each)
(a) If aH = bH, then Ha = Hb.
Let G = S3 and H = {(1), (13)}. Then
(12)H = {(12), (132)} = (132)H, but H(12) = {(12), (123)} =
6
H(132) = {(132), (23)}
(b) If Ha = Hb, then b ∈ Ha
Proof. b = eb ∈ Hb = Ha. Thus, b ∈ Ha.
(c) If aH = bH, then Ha−1 = Hb−1
Proof. We are trying to show two sets are equal, so we will
show Ha−1 ⊆ Hb−1 and Ha−1 ⊇ Hb−1 .
First, let ha−1 ∈ Ha−1 . Then ah−1 ∈ aH = bH. So,
ah−1 = bh0−1 for some h0 ∈ H. So, (ah−1 )−1 = (bh0−1 )−1 ⇒
ha−1 = h0 b−1 ∈ Hb−1 .
Next, let hb−1 ∈ Hb−1 . Then bh−1 ∈ bH = aH. So bh−1 =
ah0−1 for some h0 ∈ H. So, (bh−1 )−1 = (ah0−1 )−1 ⇒ hb−1 =
h0 a−1 ∈ Ha−1 .
(d) If aH = bH, then a2 H = b2 H
Use the G, H, a, and b from part (a). Then aH = bH but
a2 = (12)2 = (1) and b2 = (132)2 = (123). So,
a2 H = H 6= b2 H = {(123), (23)}.
(3) Fill in the 3 claims in the proof of Cayley’s Theorem, which
states that any group G is isomorphic to a group of permutations. (15pts)
4
Proof. Let G be a group. For g ∈ G define a mapping
Tg : G → G by Tg (x) = gx ∀x ∈ G. Then Tg is actually
performing left multiplication on the elements of G.
Claim 1: Tg is a permutation
Proof. To show Tg is a permutation, show Tg is 1-1 and onto.
First, suppose Tg (x) = Tg (y). Then gx = gy ⇒ x = y. So, Tg
is 1-1.
Next, let x ∈ G. Then Tg (g −1 x) = g(g −1 x) = x. So Tg is
onto.
Let G0 = {Tg | g ∈ G}.
Claim 2: G0 is a group.
Proof. Let Tg , Th ∈ G0 . Then Tg Th (x) = Tg (hx) = g(hx) =
(gh)x = Tgh (x). So, G0 is closed since Tg Th = Tgh .
Function composition is always associative so Tg (Th Tj ) = (Tg Th )Tj .
The identity in G0 is Te where e is the identity in G. This is
because Te Tg (x) = Te (gx) = e(gx) = (eg)x = gx = Tg (x). So,
Te Tg = Tg . Likewise, Tg Te = Tg .
Since Tg−1 Tg (x) = Tg−1 (gx) = g −1 (gx) = (g −1 g)x = x = Te (x),
we can say Tg−1 Tg = Te and therefore, (Tg )−1 = Tg−1 . So, G0 is
a group.
Define φ : G → G0 by φ(g) = Tg .
phism.
Claim 3: φ is an isomor-
Proof. Suppose φ(g) = φ(h). Then Tg = Th which means gx =
hx for all x ∈ G. So, let x = e then g = h and φ is 1 − 1.
Let Tg ∈ G0 . Then g ∈ G and φ(g) = Tg so φ is onto.
φ(gh) = Tgh = Tg Th = φ(g)φ(h).
Therefore, φ is an isomorphism and we have G ∼
= G0 so any
group is isomorphic to a group of permutations.
(4) Using the attached sheet, form a cube with labeled faces: 1, 2,
3, 4, 5, 6. Alternatively, you can also use a standard die. Let C
be the rotation group of the cube. Determine the 24 elements
of C by identifying each rotation as an element of S6 . This is
similar to the Dihedral worksheet done in class when the rotations of a square and pentagon were determined as elements of
S4 and S5 , respectively. (10pts)
By labeling the cube as a standard die, Face 1 will be across
from Face 6, Face 3 will be across from Face 4, and Face 5
5
will be across from Face 2. If you labeled your cube differently,
your permutations will be isomorphic to the ones listed below.
In essence, they will perform the same operation, but have a
different name.
C = {(1), (1265), (1562), (1364), (1463), (2354), (2453), (16)(25),
(16)(34), (25)(34), (124)(365), (123)(465), (154)(236), (153)(246),
(135)(264), (145)(263), (142)(356), (132)(456), (12)(56)(34),
(14)(25)(36), (13)(25)(46), (15)(26)(34), (16)(24)(35), (16)(23)(45)}
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