Download S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name

S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
Some problems are really easy, some are harder, some are repetitions.
1. Let Sn be the group of permutations on n elements {1, 2, 3, . . . , n}. Let An be the subgroup
of even permutations. Prove that An is normal subgroup of Sn .
Answer:
• One way:
◦
◦
◦
◦
|Sn | = n! and |An | = n!/2
# (left cosets of An in Sn )= [Sn : An ] = |Sn |/|An | = n!/(n!/2) = 2
# (right cosets of An in Sn )=# (left cosets of An in Sn )=2
Want to show: aAn = An a for all a ∈ Sn
- If a ∈ An then aAn = An = An a.
- If a ∈
/ An then aAn 6= An 6= An a
Sn = An ∪ aAn disjoint union of 2 distinct left cosets.
Sn = An ∪ An a disjoint union of 2 distinct right cosets.
An ∪ aAn = An ∪ An a disjoint unions implies
aAn = An a for all a ∈
/ An
◦ An is normal subgroup in Sn since aAn = An a for all a ∈ Sn
• Another way:
◦ An = {even permutations on n elements}
◦ A subgroup H in a group G is normal subgroup, if and only if gHg −1 ⊂ H. Proposition proved in class.
◦ WTS σAn σ −1 ⊂ An
Let α ∈ An . Then α is an even permutation (by definition of An ).
Then σασ −1 is even permutation for all σ ∈ Sn (by a problem from a previous quiz).
Therefore σασ −1 ∈ An
Therefore σAn σ −1 ⊂ An
Therefore An is normal subgroup in Sn
2. Let S15 be the group of permutations on 15 elements {1, 2, 3, . . . , 15}. Prove that S15 has
elements of order 56 but does not have any elements of orders 49 or 50.
Answer: Use the following facts:
(a) Order of a k-cycle is k, i.e. |α| = k for α a k-cycle.
(b) |αβ| = lcm(|α|, |β|) if α and β are disjoint permutations.
• S15 has elements of order 56: Let σ = αβ, where α is an 8-cycle and β is a 7-cycle. Then
|σ| = |αβ| = lcm(|α|, |β|) = lcm(8, 7) = 56.
• S15 does not have any elements of orders 49: In order for lcm(m, n) = 49, either m = 49
or n = 49. Therefore we would need a cycle of order 49, hence need a 49-cycle in S15 ,
which is impossible.
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
• S15 does not have any elements of orders 50: In order for lcm(m, n) = 50, either m or n
must be a multiple of 52 = 25. Therefore we would need a 25-cycle or 50-cycle in S15 ,
which is impossible.
( Factor Groups - Quotient Groups - G/H )
3. Prove that a factor group of a cyclic group is cyclic.
Answer: Recall: A group G is cyclic if it can be generated by one element, i.e. if there exists
an element a ∈ G such that G =< a > (this means that all elements of G are of the form ai
for some integer i.)
Recall: Elements of a factor group G/H are left cosets {gH | g ∈ G.
Proof: Suppose G =< a >. Let G/H be any factor group of G. WTS: G/H is cyclic.
Any element of G/H is of the form gH for some g ∈ G.
Since G is cyclic, there is an integer i such that g = ai . So gH = ai H.
ai H = (aH)i by definition of multiplication in factor groups.
Therefore gH = (aH)i for any coset gH.
So G/H is generated by aH, hence G/H is cyclic, by the definition of cyclic groups.
4. Prove that a factor group of an Abelian group is Abelian.
Answer: Recall: A group G is Abelian if ab = ba for all a, b ∈ G.
Proof: Suppose G is Abelian. Let G/H be any factor group of G. WTS: G/H is Abelian.
Let aH, bH be any elements of G/H.
Then (aH)(bH) = (ab)H by definition of multiplication in factor groups.
Then (ab)H = (ba)H since G is abelian.
(ba)H = (bH)(aH) again definition of multiplication in factor group.
Therefore (aH)(bH) = (bH)(aH).
Therefore G/H is Abelian.
5. Prove that any subgroup of an Abelian group is normal subgroup.
Answer: Recall: A subgroup H of a group G is called normal if gH = Hg for every g ∈ G.
Proof: Suppose G is Abelian. Let H be a subgroup of G. WTS: gH = Hg for all g ∈ G.
Let g ∈ G. Then gH = {gh | h ∈ H} by definition of left coset.
gh = hg for all h since G is Abelian.
Therefore {gh | h ∈ H} = {hg | h ∈ H} = Hg by definition of right coset Hg.
Therefore gH = Hg for all g ∈ G. Therefore H is normal subgroup of G.
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
6. Let H =< 5 > be the subgroup of G = Z generated by 5.
(a) List the elements of H =< 5 >.
Answer: Notice: Operation in G = Z is addition.
H = h5i = {0, ±5, ±10, . . . } = {5k | k ∈ Z}
(b) Prove that H is normal subgroup of G.
Answer: Addition in Z is commutative. So G = (Z, +) is Abelian group and by previous
problem every subgroup of an Abelian group is normal.
(c) List the elements of the factor group G/H = Z/ < 5 >.
Answer: Elements of G/H are left cosets of H in G:
0 + H = 0 + h5i = {0 + 0, 0 ± 5, 0 ± 10, . . . } = {0, ±5, ±10, . . . } = H
1 + H = 1 + h5i = {1 + 0, 1 ± 5, 1 ± 10, . . . } = {. . . , −9, −4, 1, 6, 11, . . . }
2 + H = 2 + h5i = {2 + 0, 2 ± 5, 2 ± 10, . . . } = {. . . , −8, −3, 2, 7, 12, . . . }
3 + H = 3 + h5i = {3 + 0, 3 ± 5, 3 ± 10, . . . } = {. . . , −7, −2, 3, 8, 13, . . . }
4 + H = 4 + h5i = {4 + 0, 4 ± 5, 4 ± 10, . . . } = {. . . , −6, −1, 4, 9, 14, . . . }
(d) Write the Cayley table of G/H = Z/ h5i.
G/H = Z/ h5i
0 + h5i
1 + h5i
2 + h5i
3 + h5i
4 + h5i
0 + h5i
0 + h5i
1 + h5i
2 + h5i
3 + h5i
4 + h5i
1 + h5i
1 + h5i
2 + h5i
3 + h5i
4 + h5i
0 + h5i
(e) What is the order of 2 + h5i in Z/ h5i?
Answer: |2 + h5i | = 5 as 2 + h5i ∈ Z/ h5i.
3
2 + h5i
2 + h5i
3 + h5i
4 + h5i
0 + h5i
1 + h5i
3 + h5i
3 + h5i
4 + h5i
0 + h5i
1 + h5i
2 + h5i
4 + h5i
4 + h5i
0 + h5i
1 + h5i
2 + h5i
3 + h5i
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
7. Let K = h15i be the subgroup of G = Z generated by 15.
(a) List the elements of K = h15i.
Answer: K = h15i = {15k | k ∈ Z}
(b) Prove that K is normal subgroup of G.
Proof: (Z +) is Abelian group and any subgroup of an Abelian group is normal (from 5).
(c) List the elements of the factor group G/K = Z/ h15i.
Answer: G/K = Z/ h15i = {i + h15i | 0 ≤ i ≤ 14}. (There are 15 elements.)
!!! This is just one way of expressing these cosets. Notice that there are many ways of
expressing the same coset, eg. (2+h15i) = (17+h15i) = (32+h15i) = (−13+h15i) = . . . .
(d) Write the Cayley table of G/K = Z/ h15i.
Answer: Don’t have to actually write it, but make sure that you know what it would
look like.
(e) What is the order of 3 + K = 3 + h15i in Z/ h15i?
Answer: |3 + h15i | = 5 in Z/ h15i, since
(3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) = (15 + h15i) = (0 + h15i).
(f) What is the order of 4 + K = 4 + h15i in Z/ h15i?
Answer: |4 + h15i | = 15 in Z/ h15i.
(g) What is the order of 5 + K = 5 + h15i in Z/ h15i?
Answer: |5 + h15i | = 3 in Z/ h15i.
(h) What is the order of 6 + K = 6 + h15i in Z/ h15i?
Answer: |6 + h15i | = 5 in Z/ h15i,
(i) Prove that G/K is cyclic.
Answer: Z/ h15i is generated by 1 + h15i, hence it is cyclic.
underlineAnswer 2: Z/ h15i is cyclic since it is factor of the cyclic group (Z, +) (this
group is generated by 1).
(j) Prove that G/K = Z/ h15i is isomorphic to Z15 .
Answer:
• One way - using the First Isomorphism Theorem:
– Define a group homomorphism: f : Z → Z15 :
∗ Since Z is cyclic it is enough to define homomorphism on a generator, and
extend to all other elements.
∗ Define f (1) := 1(mod15) and f (n1) := n1 (mod15) (i.e. f (n) := n (mod15)
∗ Since |1| = ∞ for 1 ∈ Z and |f (1)| = 15 for f (1) = 1 ∈ Z/ h15i we have
|f (1)| | |1|.
– Claim 1: f is onto (this is quite clear, but here is a detailed proof).
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
∗ Elements of Z15 are integers {0, 1, 2, . . . , 14}
∗ For each n ∈ {0, 1, 2, . . . , 14} = Z15 consider n ∈ Z.
∗ Then f (n) = n ∈ Z15 . Hence f is onto.
– Im(f ) = Z15 (As stated in class this is equivalent to f being onto.)
– Claim 2: Ker(f ) = h15i < Z (this is quite clear, but here is a detailed proof).
∗ Ker(f ) = {n ∈ Z | f (n) = 0 ∈ Z15 }
∗ n (mod15) = r ∈ {0, 1, . . . , 14}, where r is the remainder after dividing n by
15, i.e. n = 15k + r, for some integer k.
∗ If n ∈ Ker(f ) then f (n) = 0 and therefore n = 15k ∈ h15i < Z.
∗ Therefore Ker(f ) ⊂ h15i.
∗ h15i ⊂ Ker(f ) is clear since x ∈ h15i implies x = 15k for some integer k and
therefore f (x) = f (15k) = 15k(mod15) = 0inZ15 .
– First Isomorphism Theorem states: If f : G → G0 is a group homomorphism
then G/Ker(f ) ∼
= Im(f ).
∼
– Z/ h15i = Z15
• Another way, by defining isomorphism and checking all details of being isomorphism:
– Elements of Z/ h15i are left cosets, and operation is addition of cosets (as defined
for factor groups!)
– Elements of Z15 : {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} and addition is modulo
15.
– Define the map: f : Z/ h15i → Z15 by f (i + h15i) = ri , where ri is the remainder
after dividing i by 15.
∗ Notice that f , as defined, is a mapping from Z/ h15i to Z15 since the values of
f are the remainders after dividing by 15, hence they are non-negative integers
between 0 and 14.
∗ Prove that f is well-defined:
Suppose the same coset (i + h15i) is given by another integer j, i.e.
(i + h15i) = (j + h15i).
Then f (i + h15i) = ri ∈ Z15 and f (j + h15i) = rj ∈ Z15 .
To show that f is well defined we must show that f (i + h15i) = f (j + h15i),
i.e. WTS ri = rj ∈ Z15 .
From (i + h15i) = (j + h15i) it follows that j ∈ (i + h15i) and therefore
j = i + k15 for some integer k ∈ Z.
By definition of remainders: i = n · 15 + ri and j = m · 15 + rj , with 0 ≤
ri , rj < 15.
Therefore from j = i + k15, we have m · 15 + rj = n · 15 + ri + k15. Hence
rj −ri = (k+n−m)·15 which is an integer multiple of 15. Since the remainders
are 0 ≤ ri , rj < 15, it follows that rj = ri .
Therefore f is a well defined function.
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
∗ Let f : Z/ h15i → Z15 defined by f (i + h15i) = ri , where ri is the remainder
after dividing i by 15. Then f is ”one-to-one” function.
Proof: Suppose there are a, b ∈ Z/ h15i such that f (a) = f (b). WTS a = b.
From a, b ∈ Z/ h15i, it follows that a = i + h15i and b = j + h15i.
From f (a) = f (b) it follows f (i + h15i) = f (j + h15i), and by definition of f
it follows that ri = rj .
So i = n · 15 + ri and j = m · 15 + rj . Therefore i − j is a multiple of 15, hence
and element of the subgroup h15i.
Therefore a = i + h15i = j + h15i = b.
Therefore f is a one-to-one function.
∗ f : Z/ h15i → Z15 is onto.
Proof: Let y ∈ Z15 . WTS: there is an x ∈ Z/ h15i such that f (x) = y.
Since y ∈ Z15 , y is an integer 0 ≤ y < 15.
Let x = y + h15i ∈ Z/ h15i.
Then f (x) = f (y + h15i) = ry . Since 0 ≤ y < 15, it follows that ry = y.
Therefore f (x) = y.
Therefore f is onto.
∗ f (a + b) = f (a) + f (b)
Proof: Let a, b ∈ Z/ h15i.
From a, b ∈ Z/ h15i, it follows that a = i + h15i and b = j + h15i.
Then it follows that a + b = (i + h15i) + (j + h15i) = (i + j) + h15i
f (a + b) = (i + j) (mod15)
f (a) + f (b) = i (mod15) + j (mod15) = (i + j) (mod15)
Therefore f (a + b) = f (a) + f (b)
– Therefore f : Z/ h15i → Z15 is an isomorphism.
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
8. Let G = h6i and H = h24i be subgroups of Z. Show that H is a normal subgroup of G. Write
the cosets of H in G. Write the Cayley table for G/H.
Answer: G and H are subgroups of Z, hence operation is addition.
G = h6i = {0, ±6, ±12, ±18, ±24, ±30, ±36, ±42, ±48, . . . } = {6j | j ∈ Z}, multiples of 6.
H = h24i = {0, ±24, ±48, . . . } = {24j | j ∈ Z}, i.e. all integer multiples of 24.
• H is a normal subgroup of G.
– H is a nonempty subset of G, since elements of H are elements of G (integer multiples
of 24 are multiples of 6).
– H is closed under operation: If a, b ∈ H, then a = 24m, and b = 24n. Therefore
a + b = 24m + 24n = 24(m + n) ∈ H.
– H is closed under inverses: If a ∈ H, then a = 24m. Therefore −a = 24 · (−m),
hence −a ∈ H.
– Since Z is abelian, G is also abelian and therefore any subgroup of G is normal.
Hence H is normal in G.
• Cosets of H = h24i in G = h6i are:
H = 0 + H = 0 + h24i = {. . . , −48, −24, 0, 24, 48, . . . },
6 + H = 6 + h24i = {. . . , −42, −18, 6, 30, 54, . . . },
12 + H = 12 + h24i = {. . . , −36, −12, 12, 36, 60, . . . },
18 + H = 18 + h24i = {. . . , −30, −6, 18, 42, 66, . . . }.
• Elements of G/H are the four cosets written above and the Cayley table is:
G/H = h6i / h24i
0 + h24i
6 + h24i
12 + h24i
18 + h24i
0 + h24i 6 + h24i
0 + h24i 6 + h24i
6 + h24i 12 + h24i
12 + h24i 18 + h24i
18 + h24i 0 + h24i
12 + h24i 18 + h24i
12 + h24i 18 + h24i
18 + h24i 0 + h24i
0 + h24i 6 + h24i
6 + h24i 12 + h24i
9. Viewing h6i and h24i as subgroups of Z, prove that h6i/h24i is isomorphic to Z4 .
Proof: Elements of h6i/h24i are the 4 cosets: {(0 + h24i), (6 + h24i), (12 + h24i), (18 + h24i)}
with the above multiplication table.
|6 + h24i | = 4 since (6 + h24i) + (6 + h24i) + (6 + h24i) + (6 + h24i) = (0 + h24i).
So h6i/h24i can be generated by one element of order 4, therefore it is cyclic of order 4.
The group Z4 is also cyclic of order 4. By theorem: ”Any two cyclic groups of the same order
are isomorphic.”, it follows that h6i/h24i is isomorphic to Z4 .
7
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
10. Let h8i be the subgroup of Z48 .
(a) What is the order of the factor group Z48 /h8i?
Answer: Elements of Z48 are {0, 1, 2, 3, . . . 46, 47} and |Z48 | = 48.
Elements of h8i ⊂ Z48 are {0, 8, 16, 24, 32, 40} and | h8i | = 6.
Therefore the order of the factor group is: |Z48 /h8i | = |Z48 |/| h8i | = 48/6 = 8.
(b) What is the order of 2 + h8i in the factor group Z48 /h8i?
Answer: Elements of Z48 / h8i are the following cosets:
{(0 + h8i), (1 + h8i), (2 + h8i), (3 + h8i), (4 + h8i), (5 + h8i), (6 + h8i), (7 + h8i)}.
The order of (2 + h8i) is |(2 + h8i | = 4 since 4 is the smallest number of times (2 + h8i)
must be added to itself in order to get the identity, i.e. such that (2 + h8i) + (2 + h8i) +
(2 + h8i) + (2 + h8i) = (0 + h8i).
11. Let G = U (16) be the group of units modulo 16.
(a) What is the order of G?
Answer: G = {1, 3, 5, 7, 9, 11, 13, 15}. So |G| = 8.
We also know that |U (16)| = φ(16) = φ(24 ) = 23 (2 − 1) = 8.
(b) What is the order of 15 ∈ U (16)?
Answer: 15 · 15 = 225 ≡ 1(mod16). Therefore |15| = 2 in U (16).
(Another way: 15 · 15 = (−1)(−1) = 1 ≡ 1(mod16))
(c) Let H =< 15 > be the subgroup of U (16) generated by 15. What is the order of the
factor group U (16)/H?
Answer: |G| = 8, |H| = 2. Therefore |G/H| = |G|/|H| = 8/2 = 4
Also, in more details: H =< 15 >= {1, 15}.
Elements of G/H are the left cosets of H in G:
1H = {1, 15} = 15H,
3H = {3, 13} = 13H
5H = {5, 11} = 11H
7H = {7, 9} = 9H
(d) Make the Cayley table of the factor group U (16)/H.
Answer:
G/H
1H
3H
5H
7H
1H
1H
3H
5H
7H
3H
3H
7H
1H
5H
8
5H
5H
1H
7H
3H
7H
7H
5H
3H
1H
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
( Group Homomorphisms )
12. Let F : Z → Z be a function defined as F (x) = 10x.
(a) Prove that F is a group homomorphism.
Proof:
• F is a function F : Z → Z (already stated in the problem, so no need to prove it).
• F (a + b) = 10(a + b) by definition of F .
10(a + b) = 10a + 10b distributive property of integers.
F (a) + F (b) = 10a + 10b by definition of F .
Therefore F (a + b) = F (a) + F (b).
• Therefore F is a group homomorphism.
(b) Find Ker(F ).
Answer:
• Ker(F ) = {x ∈ Z | F (x) = 0 ∈ Z} = {x ∈ Z | 10x = 0 ∈ Z} = {0}.
• Ker(F ) = {0}
(c) Find Im(F )
Answer:
• Im(F ) = {y ∈ Z | there is an x ∈ Z such that F (x) = y} = {10x ∈ Z | x ∈ Z} = 10Z.
• Im(F ) = h10i < Z
13. Let f : Z6 → D3 be the map defined as f (1) = ρ.
(a) Find f (2)
f (2) = f (1 + 1) = f (1)f (1) = ρρ = ρ2
(b) Find f (3)
f (3) = f (1 + 1 + 1) = f (1)f (1)f (1) = ρρρ = ρ3 = e ∈ D3
(c) Find f (4)
f (4) = f (4 · 1) = f (1)4 = ρ4 = ρ
(d) Find f (5)
f (5) = f (1)5 = ρ5 = ρ2
(e) Find f (0)
f (0) = e
(f) Find Ker(f )
Ker(f ) = {0, 3} = h3i < Z6
(g) Find Im(f )
Im(f ) = {e, ρ, ρ2 } = hρi < D3
(h) Find Z6 /Ker(f )
Z6 /Ker(f ) = {Ker(f ), 1+Ker(f ), 2+Ker(f )} = {h3i , 1+h3i , 2+h3i}
9
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
(i) Describe an isomorphism Z6 /Ker(f ) → Im(f )
Define ϕ : Z6 /Ker(f ) → Im(f )
Define ϕ : Z6 / h3i → hρi < D3 by
ϕ(1+h3i) = ρ and therefore
ϕ(2+h3i) = ρ2 and
ϕ(h3i)) = ϕ(0+h3i) = ϕ(3+h3i) = ρ3 = e.
14. Let f : G → G0 be a group homomorphism. Prove: If Ker(f ) = e then f is a one-to-one map.
15. Let f : G → G0 be a group homomorphism. Prove: Ker(f ) is a normal subgroup of G.
16. If |G| = 18 and a ∈ G, what are all possible orders of a?
Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G|
• |a| = 1, 2, 3, 6, 9, 18
17. If |G| = 17 and a ∈ G, what are all possible orders of a?
Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G|
• |a| = 1, 17
18. If |G| = 23 and a ∈ G, a 6= e, what are all possible orders of a?
Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G|
• |a| = 1, 23
• |a| =6= 1 since a 6= e and e is the only element in G od order 1.
• |a| = 23
19. Let f : Z15 → Z12 be a group homomorphism defined by f (x) = 4x(mod 12).
• Find f (5).
f (5) = 4 · 5 (mod12) = 20 (mod12) = 8 ∈ Z12
• Find f (6).
f (6) = 4 · 6 (mod12) = 24 (mod12) = 0 ∈ Z12
• Find Im(f ).
Im(f ) = {4, 8, 0} = h4i < Z12
• Find Ker(f ).
Ker(f ) = {3, 6, 9, 12, 0} = h3i < Z15
• Is f onto?
No, since Im(f ) 6= Z12
10
S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
• Is f one-to-one?
No, since Ker(f ) 6= {0}.
Also, easy to see: there are 5 different elements which are all mapped to the same element.
• Is f an isomorphism?
No, f is neither one-to-one, not onto.
20. Let f : Z15 → Z25 be a group homomorphism. What are all possible orders of f (1)?
• |1| = 15 in Z15
• |f (a)| | |a| implies that |f (1)| = 1, 3, 5, 15.
• |g| | |G| implies that f (1)|25 and therefore f (1) = 1, 5, 25
• Therefore: |f (1)| = 1, 5.
21. Describe all possible group homomorphisms Z15 → Z25 .
• Since Z15 is cyclic, it is enough to define homomorphism on 1 ∈ Z15 , and extend to other
elements.
• The only possibilities for the image of 1 ∈ Z15 are elements of order 1 or 5.
• Case 1: |f (1)| = 1. This implies:
f (1) = 0 and therefore for all x ∈ Z15 , f (x) = 0 ∈ Z25 .
• Case 2: |f (1)| = 5. This implies:
f (1) = 5, 10, 15, 20, i.e. all integers 1 ≤ i ≤ 25 such that gcd(i, 25) = gcd(5, 25) = 5.
So, there are 4 such homomorphisms:
f1 (1) = 5 and therefore for all x ∈ Z15 , f1 (x) = 5x ∈ Z25 ,
f2 (1) = 10 and therefore for all x ∈ Z15 , f2 (x) = 10x ∈ Z25 ,
f3 (1) = 15 and therefore for all x ∈ Z15 , f3 (x) = 15x ∈ Z25 ,
f4 (1) = 20 and therefore for all x ∈ Z15 , f4 (x) = 20x ∈ Z25 .
22. Describe all possible group isomorphisms Z15 → Z25 .
• There are no isomorphisms, since the groups have different orders:
|Z15 | = 15 and |Z25 | = 25 and therefore cannot be isomorphic.
23. Describe all possible group isomorphisms Z25 → Z25 .
• Since Z25 is cyclic, it is enough to define homomorphism on 1 ∈ Z25 , and extend to other
elements.
• In order for the homomorphism to be isomorphism the order of f (1) must be 25.
• The elements of order 25 in Z25 are 1 ≤ i ≤ 25 such that gcd(i, 25) = gcd(1, 25) = 1.
• The elements of order 25 in Z25 are {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}
• Also the number of elements in Z25 relatively prime to 25 is equal to Euler function
ϕ(25).
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
• ϕ(25) = ϕ(52 ) = 5(5 − 1) = 20. Therefore, there are 20 isomorphisms Z25 → Z25 :
f1 (x) = 1x, f2 (x) = 2x, f3 (x) = 3x, f4 (x) = 4x, f5 (x) = 6x, f6 (x) = 7x,
f7 (x) = 8x, f8 (x) = 9x, f9 (x) = 11x, f10 (x) = 12x, f11 (x) = 13x, f12 (x) = 14x,
f13 (x) = 16x, f14 (x) = 17x, f15 (x) = 18x, f16 (x) = 19x, f17 (x) = 21x, f18 (x) = 22x,
f19 (x) = 23x, f20 (x) = 24x
24. How many group isomorphisms Z75 → Z75 .
• (# group isomorphism Z75 → Z75 )= (# generators of Z75 )=
(# of integers i, such that 1 ≤ i ≤ 75 and gcd(i, 75) = 1) =
ϕ(75) = ϕ(3 · 52 ) = ϕ(3)ϕ(52 ) = (3 − 1)(5)(5 − 1) = (2)(5)(4) = 40
• There are 40 group isomorphisms Z75 → Z75 .
25. Prove that there is no group homomorphism f : Z10 → Z25 such that f (1) = 3.
• |1| = 10 in Z10 . Therefore |f (1)| = 1, 2, 5, 10
• gcd(3, 25) = 1 implies that |3| = 25 in Z25 .
• Therefore f (1) 6= 3.
26. Prove that there is exactly one group homomorphism f : Z10 → Z49 .
• |1| = 10 in Z10 . Therefore |f (1)| = 1, 2, 5, 10
• f (1) ∈ Z49 implies |f (1)| = 1, 7, 49
• |f (1)| = 1 form the above two statements.
• f (x) = 0 ∈ Z49 for all x ∈ Z10 .
27. Prove that there is exactly one group homomorphism f : Z11 → S6 .
• |1| = 11 in Z11 . Therefore |f (1)| = 1, 11
• |S6 | = 6! implies that order of elements of S6 must divide 6!.
• Since 11 does not divide 6!, the order of f (1) must be 1.
• f (x) = e ∈ S6 for all x ∈ Z11
28. Describe all group homomorphisms Z9 → S3 .
In each case describe Image and Kernel of the homomorphism.
• |1| = 9 in Z9 . Therefore |f (1)| = 1, 3, 9
• S3 has elements of order 1,2,3.
• The possible orders for f (1) are 1,3, and the possible functions:
– f1 : Z9 → S3 is defined as f1 (i) = (1)
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
∗ Im(f1 ) = {e} = {(1)} = h(1)i < S3 (These are just different forms of writing
the same subgroup.)
∗ Ker(f1 ) = Z9
– f2 (1) = (123), and extend to all i ∈ Z9 :
∗ f2 : Z9 → S3 is defined as f2 (i) = (123)i
∗ Computtaions of f2 (i): f2 (2) = (123)2 = (132), f2 (3) = (1), f2 (4) = (123)4 = (123),
f2 (5) = (123)5 = (132), f2 (6) = (123)6 = (1), f2 (7) = (123)7 = (123),
f2 (8) = (123)8 = (132), f2 (0) = f2 (9) = (123)9 = (1),
∗ Im(f2 ) = {(123), (132), (1)} = h(123)i < S3
∗ Ker(f2 ) = {3, 6, 0} = h3i < Z9
– f3 (1) = (132), and extend to all i ∈ Z9 :
∗ f3 : Z9 → S3 is defined as f3 (i) = (132)i
∗ Im(f3 ) = {(132), (123), (1)} = h(132)i < S3
∗ Ker(f3 ) = {3, 6, 0} = h3i < Z9
29. Describe all group homomorphisms Z9 → S4 .
In each case describe Image and Kernel of the homomorphism.
• Possible orders for f (1) are 1 or 3.
• f1 (x) = e, for all x ∈ Z9
Ker(f1 ) = Z9
Im(f1 ) = h(123)i
• f2 (x) = (123)x , for all x ∈ Z9
Ker(f2 ) = h3i < Z9
Im(f2 ) = h(123)i
• f3 (x) = (132)x , for all x ∈ Z9
Ker(f3 ) = h3i < Z9
Im(f3 ) = h(132)i
• f4 (x) = (124)x , for all x ∈ Z9
Ker(f4 ) = h3i < Z9
Im(f4 ) = h(124)i
• f5 (x) = (142)x , for all x ∈ Z9
Ker(f5 ) = h3i < Z9
Im(f5 ) = h(142)i
• f6 (x) = (134)x , for all x ∈ Z9
Ker(f6 ) = h3i < Z9
Im(f6 ) = h(134)i
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
• f7 (x) = (143)x , for all x ∈ Z9
Ker(f7 ) = h3i < Z9
Im(f7 ) = h(143)i
• f8 (x) = (234)x , for all x ∈ Z9
Ker(f8 ) = h3i < Z9
Im(f8 ) = h(234)i
• f9 (x) = (243)x , for all x ∈ Z9
Ker(f9 ) = h3i < Z9
Im(f9 ) = h(243)i
30. Describe all group homomorphisms Z9 → U (10).
In each case describe Image and Kernel of the homomorphism.
• Order of 1 ∈ Z9 is 9. Therefore |f (1)| = 1, 3, 9.
• U (10) = {1, 3, 7, 9}, therefore |U (10)| = 4. So |f (1)| = 1, 2, 4
• Therefore |f (1)| = 1. Hence f (1) = 1 ∈ U (10). Extend to all i ∈ Z9 .
• f (i · 1) = 1i = 1 ∈ U (10) for each i ∈ Z9 .
• Im(f ) = {1} = h1i < U (10).
• Ker(f ) = Z9 .
31. Describe all group homomorphisms Z9 → U (9).
In each case describe Image and Kernel of the homomorphism.
• Order of 1 ∈ Z9 is 9. Therefore |f (1)| = 1, 3, 9.
• U (9) = {1, 2, 4, 5, 7, 8}, therefore |U (9)| = 6. So |f (1)| = 1, 2, 3, 6
• Therefore |f (1)| = 1, 3.
• Case 1. |f (1)| = 1.
– Then f (1) = 1 ∈ U (9). Extend to all i ∈ Z9 in order to get: f : Z9 → U (9)
– f (i · 1) = 1i = 1 ∈ U (9) for each i ∈ Z9 .
– Im(f ) = {1} = h1i < U (9).
– Ker(f ) = Z9 .
• Case 2. |f (1)| = 3.
– First find orders of each of the elements in U (9):
|1| = 1 Since order of the identity in any group is always 1.
|2| = 6 Since 21 = 2, 22 = 4, 23 = 8, 24 = 7, 25 = 5, 26 = 1 ∈ U (9)
|4| = 3 Since 41 = 4, 42 = 7, 43 = 1 ∈ U (9)
|5| = 6 Since 51 = 5, 52 = 7, 53 = 8, 54 = 4, 55 = 2, 56 = 1 ∈ U (9)
|7| = 3 Since 71 = 7, 72 = 4, 73 = 1 ∈ U (9)
|8| = 2 Since 81 = 8, 82 = 1 ∈ U (9)
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S11MTH 3175 Group Theory (Prof.Todorov)
Quiz 5 (Practice)
Name:
– So, there are two more homomorphisms: f2 (1) = 4 and f3 (1) = 7
– f2 (1) = 4 ∈ U (9).
∗ Extend to all i ∈ Z9 in order to get: f2 : Z9 → U (9)
∗ f2 (i · 1) = 4i ∈ U (9) for each i ∈ Z9 .
∗ f2 (1) = 4, f2 (2) = 42 = 7, f2 (3) = 43 = 1, f2 (4) = 44 = 4, f2 (5) = 45 = 7,
f2 (6) = 46 = 1, f2 (7) = 47 = 4, f2 (8) = 48 = 7, f2 (0) == f2 (9) = 49 = 1
∗ Im(f2 ) = {4, 7, 1} = h4i < U (9).
∗ Ker(f2 ) = {3, 6, 0} = h3i < Z9 .
– f3 (1) = 7 ∈ U (9).
∗ Extend to all i ∈ Z9 in order to get: f3 : Z9 → U (9)
∗ f3 (i · 1) = 7i ∈ U (9) for each i ∈ Z9 .
∗ f3 (1) = 7, f3 (2) = 72 = 4, f3 (3) = 73 = 1, f3 (4) = 74 = 7, f3 (5) = 75 = 4,
f3 (6) = 76 = 1, f3 (7) = 77 = 7, f3 (8) = 78 = 4, f3 (0) == f3 (9) = 79 = 1
∗ Im(f3 ) = {4, 7, 1} = h4i < U (9).
∗ Ker(f3 ) = {3, 6, 0} = h3i < Z9 .
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