Indian Institute of Technology Guwahati
Solution to Tutorial Sheet-2
July-Nov 2015 semester
MA 521, Modern Algebra
Instructor: Shyamashree Upadhyay
Ans 1:
Since 0(𝑎) = 12 & 0(𝑏) = 22, therefore
| < 𝑎 > | = 12 &| < 𝑏 > | = 22
Now < 𝑎 >∩< 𝑏 > is a subgroup of both < 𝑎 > and < 𝑏 >
Therefore,0(< 𝑎 >∩< 𝑏 >) must divide | < 𝑎 > | = 12 as well as
| < 𝑏 > | = 22 [Lagrange’s theorem ]
The only common divisors of 12 & 22 are 1 and 2.
Since < 𝑎 >∩< 𝑏 >≠ {𝑒}, therefore it follows that
0(< 𝑎 >∩< 𝑏 >) = 2
Ans 2:
If 𝑎 + 𝑖𝑏 and 𝑐 + 𝑖𝑑 ∈ 𝐻, then (𝑎 + 𝑖𝑏)(𝑐 + 𝑖𝑑)−1 = (𝑎𝑐 + 𝑏𝑑) + 𝑖(𝑏𝑐 − 𝑎𝑑)
𝑎𝑛𝑑(𝑎𝑐 + 𝑏𝑑)2 + (𝑏𝑐 − 𝑎𝑑)2 = 1,
So that 𝐻 is a subgroup. 𝐻 is the unit circle in the complex plane ( geometrically).
Ans 3:
𝐻 is not a subgroup
1
2
[
0
0
0
] = [ 2 1 ] is not in 𝐻
2
0
2
Ans 4:
Let 𝑆 = {𝑥 ∈⊄∗| 𝑥 𝑛 = 1}
Then 𝑆 = {𝑐𝑜𝑠
2𝜋𝑘
𝑛
+ 𝑖 sin
2𝜋𝑘
|0
𝑛
≤ 𝑘 < 𝑛}[ Easy to check!]
Now let
𝑎 and 𝑏 belong to 𝑆, say
2𝜋𝑘
2𝜋𝑘
+ 𝑖 sin 𝑛
𝑛
2𝜋𝑚
2𝜋𝑚
cos 𝑛 + 𝑖 sin 𝑛 for some integers 𝑘, 𝑚
2𝜋(𝑛−𝑚)
2∏(𝑛−𝑚)
+ 𝑖 sin
[Easy to check!]
𝑛
𝑛
𝑎 = cos
and 𝑏 =
Then 𝑏 −1 = cos
such that 0 ≤ 𝑘, 𝑚 < 𝑛.
And
𝑎𝑏 −1 = cos
2𝜋
(𝑘
𝑛
+ 𝑛 − 𝑚) + 𝑖 sin
2𝜋
(𝑘
𝑛
+ 𝑛 − 𝑚)
Let 𝑃&𝑄 be the quotient and remainder upon dividing 𝑘 + 𝑛 − 𝑚 by 𝑛
It is then easy to check that cos
2𝜋
(𝑘
𝑛
+ 𝑛 − 𝑚) = cos
2∏𝑄
𝑛
and sin
2𝜋
(𝑘
𝑛
+ 𝑛 − 𝑚)sin
2𝜋𝑄
𝑛
Where 0 ≤ 𝑄 < 𝑛.
Hence 𝑎𝑏 ∈ 𝑆 for any 𝑎, 𝑏 ∈ 𝑆. (proved)
Therefore, by one-step subgroup test, 𝑆 is 𝑎 subgroup of ⊄⋆
−1
Ans 5:
∁(𝑎) = {𝑥 ∈ 𝐺|𝑥𝑎 = 𝑎𝑥}
Let 𝑥, 𝑦 ∈ ∁(𝑎)
Then 𝑥𝑎 = 𝑎𝑥 and 𝑦𝑎 = 𝑎𝑦
Since 𝑦𝑎 = 𝑎𝑦, we have 𝑎 = 𝑦 −1 𝑎𝑦 ⟹ 𝑎𝑦 −1 = 𝑦 −1 𝑎
Now 𝑥𝑦 −1 𝑎 = 𝑥(𝑎𝑦 −1 ) = (𝑥𝑎)𝑦 −1 = 𝑎𝑥𝑦 −1 (proved)
Hence 𝑥𝑦 −1 ∈ ∁(𝑎)∀𝑥, 𝑦 ∈ ∁(𝑎)
Therefore, by one-step subgroup test,∁(𝑎) is a subgroup of 𝐺.
Ans 6:
The center Z(𝐺) of a group 𝐺 is defined as
Z(𝐺) = {𝑥 ∈ 𝐺|𝑥𝑎 = 𝑎𝑥∀𝑎 ∈ 𝐺}
Proof of the fact that 𝒵(𝐺) ≤ 𝐺 is similar to the proof in 𝑄5
Ans 7:
(𝐢)
0
∁ ([
1
𝑎
= {[
𝑐
𝑎
= {[
𝑐
1
𝑎 𝑏
𝑎 𝑏 0
]) = {[
] ∈ 𝐺𝐿(2,1𝑅) |[
][
0
𝑐 𝑑
𝑐 𝑑 1
𝑏
𝑏 𝑎
𝑐 𝑑
] ∈ 𝐺𝐿 (2,1𝑅 | [
]=[
]}
𝑑
𝑑 𝑐
𝑎 𝑏
𝑏
] ∈ 𝐺𝐿(2,1𝑅)|𝑏 = 𝑐, 𝑎 = 𝑑}
𝑑
1
0 1 𝑎
]=[
][
0
1 0 𝑐
𝑏
]}
𝑑
(ii)
𝑥 𝑦 𝑎 𝑏
𝑥 𝑦
𝑏
𝑎 𝑏 𝑥 𝑦
] ∈ 𝐺𝐿(2, 𝑅)| [
][
]=[
][
]∀[
] ∈ 𝐺𝐿(2,1𝑅)}
𝑢 𝑣 𝑐 𝑑
𝑢 𝑣
𝑑
𝑐 𝑑 𝑢 𝑣
𝑎𝑥 + 𝑏𝑢 𝑎𝑦 + 𝑏𝑣
𝑥 𝑦
𝑎𝑥 + 𝑐𝑦 𝑏𝑥 + 𝑑𝑦
𝑎 𝑏
= {[
] ∈ 𝐺𝐿(2, 𝑅)| [
]=[
]∀[
] ∈ 𝐺𝐿(2,1𝑅)}
𝑢
𝑣
𝑐𝑥
+
𝑑𝑢
𝑐𝑦
+
𝑑𝑣
𝑐 𝑑
𝑎𝑢 + 𝑐𝑣 𝑏𝑢 + 𝑑𝑣
Now 𝑎𝑥 + 𝑏𝑢 = 𝑎𝑥 + 𝑐𝑦 ⟹ 𝑏𝑢 = 𝑐𝑦
𝑥 𝑦
Since our choice of [
] ∈ 𝐺𝐿(2,1𝑅)is arbitrary, we see that 𝑏𝑢 = 𝑐𝑦 must hold true for any choice of
𝑢 𝑣
𝑥 𝑦
[
]
𝑢 𝑣
𝑥 𝑦
1 1
Let us take [
]=[
]
𝑢 𝑣
0 1
Then 𝑏𝑢 = 𝑐𝑦 ⟹ 0 = 𝑐
𝑥 𝑦
1 0
Similarly, taking [
]=[
], we get 𝑏 = 0
𝑢 𝑣
1 1
Together ,we get 𝑏 = 𝑐 = 0
𝑎
𝑍(𝐺) = {[
𝑐
𝑥 𝑦
Now since our choice of [
] ∈ 𝐺𝐿(2, 𝑅) is arbitrary, therefore 𝑎𝑦 + 𝑏𝑣 = 𝑏𝑥 + 𝑑𝑦 and 𝑏 = 𝑐 = 0
𝑢 𝑣
together give 𝑎 = 𝑑
𝑎 0
Therefore Z(𝐺) = {(
) |𝑎 ∈ 1𝑅, 𝑎 ≠ 0}
0 𝑎
It is easy to check that for any 𝐶 ∈ 𝐺𝐿(2, 𝑅),and for any 𝑎 ≠ 0
𝑎 0
1 0
𝑎 0
𝐶(
) = 𝑎𝐶 (
) = 𝑎(
) 𝐶 (Proved)
0 𝑎
0 1
0 𝑎
Ans 8:
Let 𝑚 = 𝑜(𝑥 −1 𝑎𝑥)
Then 𝑚 is the smallest positive integer such that (𝑥 −1 𝑎𝑥)𝑚 = 𝑒
That is, 𝑥 −1 𝑎𝑚 𝑥 = 𝑒
Clearly, (𝑥 −1 𝑎𝑥)10 = 𝑒 ⟹ 𝑜(𝑥 −1 𝑎𝑥) divides 10
So, 𝑚 = 10 is one possibility of 𝑜(𝑥 −1 𝑎𝑥)
If 𝑚 < 10, then 𝑥 −1 𝑎𝑚 𝑥 = 𝑒 ⟹ 𝑎𝑚 = 𝑒, which is a contradiction since 𝑜(𝑎) = 10
Hence 𝑚 = 10.
Ans 9:
Let 𝑝 be an odd prime. Let 𝐺 be a group with exactly 𝑝 elements of order 𝑝. let g be one of
those elements. Then the cyclic subgroup generated by 𝑔 has 𝑝 − 1 generators, nameles, 𝑔, 𝑔2 ,
. . . 𝑔𝑝−1 each of which has order 𝑝.
Thus there exists one more element in 𝐺 of order 𝑝, say 𝑥, such the 𝑥 is not a power of 𝑔
But then 𝑥 , 𝑥 2 , . . . . 𝑥 𝑝−1 all have order 𝑝, and all generate the same subgroup < 𝑥 >
Observe that the sets {𝑔, 𝑔2 , . . . , 𝑔𝑝−1 } and {𝑥, 𝑥 2 , . . . , 𝑥 𝑝−1 } are disjoint
So, 𝐺 must have at least 2 𝑝 − 2 many elements of order 𝑝.
Since 𝑝 is an odd prime. 2 𝑝 − 2 > 𝑝, a contradiction.
Ans 10:
Let 𝑔 ∈ 𝐺 be such that 𝑔 ≠ 𝑒. since 𝐺 is finite, 𝑜(𝑔) = 𝑛 > 1. Let $p$ be a prime factor of n and
let n=mp. Observe that o(𝑔𝑚 ) = 𝑝. (proved).
Ans 11:
It may not be true that 𝑜(𝑎) = 𝑜(𝑏)
Because let 𝐺 = 𝑍2 = {0 ,1}
Let 𝑎 = 0, 𝑏 = 1
2
Then 𝑎 = 0, 𝑏 2 = 0
So, 𝑜(𝑎2 ) = 𝑜(𝑏 2 ) = 1
But 𝑜(𝑎) = 1 ≠ 2 = 𝑜(𝑏).
Ans 12.
In view of the two-step subgroup test, we only need to show that 𝑎−1 ∈ 𝑆 whenever 𝑎 ∈ 𝑆.
If 𝑎 = 𝑒, then 𝑎−1 = 𝑎 and we are done
If 𝑎 ≠ 𝑒, then consider the sequence 𝑎, 𝑎2 , 𝑎3 ,. . . . .
Since 𝑆 is finite and closed under the group multiplication, we get that all positive powers of 𝑎
are in 𝑆 and not all of these positive powers of 𝑎 are distinct.
Say 𝑎𝑖 = 𝑎 𝑗 for some 𝑖 > 𝑗
Then 𝑎𝑖−𝑗 = 𝑒
Since 𝑎 ≠ 𝑒, we have 𝑖 − 𝑗 > 1.
Thus 𝑎𝑖−𝑗 = 𝑎. 𝑎𝑖−𝑗−1 = 𝑒
Therefore, 𝑎𝑖−𝑗−1 = 𝑎−1
But 𝑖 − 𝑗 − 1 ≥ 1 ⟹ 𝑎𝑖−𝑗−1 ∈ 𝑆 and we are done.
Ans 13:
(i)
Easy to prove
(ii)
Let 𝐺 = 𝒁, 𝐻 = 2𝐙, 𝐾 = 3𝐙
Then 𝐻 ∪ 𝐾 is not a subgroup of 𝐺 because
3,2 ∈ 𝐻 ∪ 𝐾 but 5=2+3 ∉ 𝐻 ∪ 𝐾
(iii)
⋈
Let 𝑎, 𝑏 ∈ ⋃ 𝐻𝑖 . Then 𝑎 ∈ 𝐻𝑛 for some 𝑛 and 𝑏 ∈ 𝐻𝑚 for some 𝑚
𝑖=1
Let 𝑁 = max (𝑛 , 𝑚)
Since 𝐻1 ⊆ 𝐻3 ⊆ . . . . . is a chain , it now follows 𝑎 , 𝑏 ∈ 𝐻𝑁
Now 𝐻𝑁 is a subgroup ⟹ 𝑎𝑏 −1 ∈ 𝐻𝑁
⋈
𝑎𝑏
−1
∈ ⋃ 𝐻𝑖
𝑖=1
Hence by one − step subgroup test, we are done
END