Homework #5
Due: October 13, 2010
1. Decide whether each of the following sets forms a group with respect to the given
operation. If it does form a group, give the identity element and the inverse of each
element. If it does not form a group, identify which of the group axioms fail.
{𝑚
}
(a) ℚ∗ =
: 𝑚, 𝑛 ∈ ℤ, 𝑚 ∕= 0, 𝑛 ∕= 0 with the operation of multiplication.
𝑛
Answer: Group. Identity = 1; (𝑚/𝑛)−1 = 𝑛/𝑚.
(b) {−1, 1} ⊂ ℤ with the operation of multiplication.
Answer: Group. Identity = 1; Each element is its own inverse.
(c) ℤ with the operation of subtraction.
Answer: Not a group. Subtraction is not associative: 𝑎 − (𝑏 − 𝑐) ∕= (𝑎 − 𝑏) − 𝑐.
{[
]
}
1 𝑛
(d) 𝐺 =
: 𝑛 ∈ ℤ with the operation of matrix multiplication.
0 1
[
][
]
1 𝑛 1 𝑚
Answer: Group. This set is closed under matrix multiplication
=
0 1 0 1
[
]
[
]−1 [
]
1 𝑛+𝑚
1 𝑛
1 −𝑛
and inverses
=
. The identity is the identity ma0
1
0 1
0 1
trix.
(e) The set 𝐹 of all functions from {1, 2, 3} to itself with the operation of composition
of functions.
Answer: Not a group. Not every function has an inverse. For example, the
constant function 𝑓 (𝑎) = 1 for all 𝑎 does not have an inverse function under
composition.
2. (a) Find elements 𝑎, 𝑏, and 𝑐 of the group 𝑆3 where 𝑎𝑏 = 𝑏𝑐, but 𝑎 ∕= 𝑐.
(
)
(
)
(
)
(
)
▶ Solution. Let 𝜎 = 12 23 31 , 𝑎 = 11 23 32 , 𝑏 = 13 22 31 , and 𝑐 = 12 21 33 . Then
𝑎𝑏 = 𝜎 = 𝑏𝑐 but 𝑎 ∕= 𝑐.
◀
(b) Find elements 𝑎 and 𝑏 of the group 𝑆3 such that (𝑎𝑏)−1 ∕= 𝑎−1 𝑏−1 .
▶ Solution. The same elements 𝑎 and 𝑏 from part (a) work.
◀
(c) Find elements 𝑎 and 𝑏 of the group 𝑆3 such that (𝑎𝑏)2 ∕= 𝑎2 𝑏2 .
▶ Solution. The same elements 𝑎 and 𝑏 from part (a) work.
◀
3. Suppose that 𝑄 is the group defined by the following multiplication table. That is,
the multiplication 𝑥𝑦 is determined by finding the element which is in the row starting
with 𝑥 and the column starting with 𝑦. For example, 𝑏𝑑 = 𝑓 . We are denoting the
Math 4023
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Homework #5
Due: October 13, 2010
identity of 𝑄 by the symbol 1.
⋅
1
𝑎
𝑏
𝑐
𝑑
𝑒
𝑓
𝑔
1
1
𝑎
𝑏
𝑐
𝑑
𝑒
𝑓
𝑔
𝑎
𝑎
1
𝑐
𝑏
𝑒
𝑑
𝑔
𝑓
𝑏
𝑏
𝑐
𝑎
1
𝑔
𝑓
𝑑
𝑒
𝑐
𝑐
𝑏
1
𝑎
𝑓
𝑔
𝑒
𝑑
𝑑
𝑑
𝑒
𝑓
𝑔
𝑎
1
𝑐
𝑏
𝑒
𝑒
𝑑
𝑔
𝑓
1
𝑎
𝑏
𝑐
𝑓
𝑓
𝑔
𝑒
𝑑
𝑏
𝑐
𝑎
1
𝑔
𝑔
𝑓
𝑑
𝑒
𝑐
𝑏
1
𝑎
(a) Does the element 𝑎 commute with all elements of 𝑄?
▶ Solution. Yes. The row and column containing 𝑎 match item by item. Thus,
𝑎𝑥 = 𝑥𝑎 for all 𝑥 ∈ 𝑄.
◀
(b) Find the element 𝑥 ∈ 𝑄 such that 𝑑𝑥 = 𝑏. For this element 𝑥, what is 𝑥𝑑?
▶ Solution. From the table, look for 𝑏 in the row headed by 𝑑. It occurs in the
column headed by 𝑓 . Thus 𝑑𝑓 = 𝑏 so 𝑥 = 𝑓 . 𝑥𝑑 = 𝑓 𝑑 = 𝑐.
◀
(c) Find all 𝑥 ∈ 𝑄 that are solutions of the equation 𝑥3 = 𝑑.
▶ Solution. 13 = 1 and 𝑎3 = 𝑎. For all other 𝑥, 𝑥2 = 𝑎 so 𝑥3 = 𝑥2 𝑥 = 𝑎𝑥.
𝑥𝑎 = 𝑑 has the solution 𝑥 = 𝑒.
◀
(d) Which elements of 𝑄 have order 2?
▶ Solution. Only 𝑎 has order 2. It is the only nonidentity element whose square
is 1.
◀
(e) Give two different elements of 𝑄 which have order 4.
▶ Solution. The elements 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 , and 𝑔 all have order 4, since for each of
these elements 𝑥4 = 1 but 𝑥2 = 𝑎 ∕= 1 and 𝑥3 = 𝑎𝑥 ∕= 1.
◀
(f) Verify that 𝐻 = {1, 𝑎} is a subgroup of 𝑄, but 𝐾 = {1, 𝑓 } is not a subgroup of
𝑄.
▶ Solution. The easiest thing is to compute the multiplication tables for 𝐻 and
𝐾:
𝐻
𝐾
⋅ 1 𝑎
⋅ 1 𝑓
1 1 𝑎
1 1 𝑓
𝑎 𝑎 1
𝑓 𝑓 𝑎
◀
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Homework #5
Due: October 13, 2010
(g) List all of the left cosets of 𝐻 in 𝑄.
▶ Solution. The left cosets are 𝐻 = {1, 𝑎}, 𝑏𝐻 = {𝑏, 𝑐}, 𝑑𝐻 = {𝑑, 𝑒} and
𝑓 𝐻 = {𝑓, 𝑔}.
◀
4. For each group in the following list, find the order of the group and the order of each
element in the group. What relation do you see between the orders of the elements of
a group and the order of the group? Recall that ℤ𝑛 denotes the group of congruence
classes mod 𝑛 with group operation addition of congruence classes, while ℤ∗𝑛 is the
group of congruence classes mod 𝑛 that have a multiplicative inverse mod 𝑛, with he
group operation being multiplication of congruence classes.
(a) ℤ12 ,
(b) ℤ∗10 ,
(c) ℤ∗12 ,
(d) 𝐷4 .
▶ Solution. (a) ∣ℤ12 ∣ = 12, 𝑜(0) = 1, 𝑜(1) = 12, 𝑜(2) = 6, 𝑜(3) = 4, 𝑜(4) = 3,
𝑜(5) = 12, 𝑜(6) = 2, 𝑜(7) = 12, 𝑜(8) = 3, 𝑜(9) = 4, 𝑜(10) = 6, and 𝑜(11) = 12. All
these are calculated from the formula: If 𝑜(𝑔) = 𝑛, then 𝑜(𝑔 𝑘 ) = 𝑛/(𝑛, 𝑘). For an
additive group this formula is 𝑜(𝑘𝑔) = 𝑛/(𝑛, 𝑘). Apply this with 𝑔 = 1 ∈ ℤ12 .
∗
(b) ∣∣𝑍10
∣ = 𝜙(10) = 𝜙(2)𝜙(5) = 4. In fact, ℤ∗10 = {1, 3, 7, 9}. This group is cyclic
with generator 3 since 32 = 9, 33 = 7, 34 = 1. Thus, 𝑜(1) = 1, 𝑜(3) = 4,
𝑜(7) = 𝑜(33 ) = 4/(4, 3) = 4, and 𝑜(9) = 𝑜(32 ) = 2.
(c) ℤ∗12 = {1, 5, 7, 11} so ∣ℤ∗12 ∣ = 4. The square of every element is 1, so 𝑜(1) = 1,
𝑜(5) = 2, 𝑜(7) = 2, and 𝑜(11) = 2.
(d) Referring to the multiplication table on Page 59 of the text, ∣𝐷4 ∣ = 8 and 𝑜(𝜖) = 1,
𝑜(𝛼) = 𝑜(𝛼3 ) = 3, and the other 5 elements have order 2.
◀
5. Show that the group ℤ∗14 is cyclic by showing that ℤ∗14 = (3) and ℤ∗14 = (5). Is
ℤ∗14 = (11)?
▶ Solution. ℤ∗14 = {1, 3, 5, 9, 11, 13} so
modulo 14 are:
1
𝑘
𝑘
3 (mod 14) 3
5𝑘 (mod 14) 5
11𝑘 (mod 14) 11
∣ℤ∗14 ∣ = 6. The powers of 3, 5, and 11
2 3 4
9 13 11
11 13 9
9 1 11
5
5
3
9
6
1
1
1
From this table it is clear that ℤ∗14 = (3) = (5) but (11) = {1, 11, 9} ∕= ℤ∗14 .
◀
∗
6. Show that ℤ∗12 ∕= (𝑘) for any 𝑘 ∈ 𝑍12
. Conclude that ℤ∗12 is not cyclic.
▶ Solution. The square of every element of ℤ∗12 is 1, so for every 𝑘, (𝑘) = {1, 𝑘} ∕= ℤ∗12 .
∗
◀
it follows that ℤ∗12 is not cyclic.
Since ℤ∗12 ∕= (𝑘) for any 𝑘 ∈ 𝑍12
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