Download Quiz 4 Practice (Some Solutions)

S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
1. Consider external direct product: Z6 ⊕ Z15 .
(a) What is the order of (2, 3) ∈ Z6 ⊕ Z15 ?
Answer:
• Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
• |(2, 3)| = lcm(|2|, |3|) = lcm(3, 5) = 15
since |2| = 6/2 = 3 in Z6 and |3| = 15/3 = 5 in Z15 .
(b) What is the order of (2, 12) ∈ Z6 ⊕ Z15 ?
Answer:
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Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
|2| = 6/2 = 3 in Z6
|12| = |gcd(12, 15)| = |3| = 15/3 = 5 in Z15 .
|(2, 12)| = lcm(|2|, |12|) = lcm(|2|, |3|) = lcm(3, 5) = 15.
(c) What is the order of (4, 12) ∈ Z6 ⊕ Z15 ?
Answer:
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Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
|4| = |gcd(4, 6)| = |2| = 6/2 = 3 in Z6 .
|12| = |gcd(12, 15)| = |3| = 15/3 = 5 in Z15 .
|(4, 12)| = lcm(|4|, |12|) = lcm(|2|, |3|) = lcm(3, 5) = 15.
(d) What is all possible orders of elements (a, b) ∈ Z6 ⊕ Z15 ?
Answer:
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Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
Orders of a ∈ Z6 must divide 6, hence |a| = 1, 2, 3, 6.
Orders of b ∈ Z15 must divide 15, hence |b| = 1, 3, 5, 15.
The only possibilities for the orders |(a, b)| are:
lcm(1, 1) = 1, lcm(1, 3) = 3, lcm(1, 5) = 5, lcm(1, 15) = 15,
lcm(2, 1) = 2, lcm(2, 3) = 6, lcm(2, 5) = 10, lcm(2, 15) = 30,
lcm(3, 1) = 3, lcm(3, 3) = 3, lcm(3, 5) = 15, lcm(3, 15) = 15,
lcm(6, 1) = 6, lcm(6, 3) = 6, lcm(6, 5) = 30, lcm(6, 15) = 30.
• Possible orders of elements (a, b) ∈ Z6 ⊕ Z15 are: 1,2,3,5,6,10,15,30.
(e) Find all elements (a, b) ∈ Z6 ⊕ Z15 of order 2.
Answer:
• Find elements (a, b) ∈ Z6 ⊕ Z15 so that |a| = 2 and |b| = 1.
• The only possibility is (3, 0).
(f) Find all elements (a, b) ∈ Z6 ⊕ Z15 of order 3.
Answer:
• Find elements (a, b) ∈ Z6 ⊕ Z15 so that:
|a| = 3, |b| = 1, or |a| = 1, |b| = 3, or |a| = 3, |b| = 3.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
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|a| = 3 in Z6 for a = 2, 4.
|b| = 3 in Z15 for b = 5, 10.
|a| = 3, |b| = 1 implies (a, b) = (2, 0), (4, 0)
|a| = 1, |b| = 3 implies (a, b) = (0, 5), (0, 10)
|a| = 3, |b| = 3 implies (a, b) = (2, 5), (4, 5), (2, 10), (4, 10)
• Elements of order 3 in Z6 ⊕Z15 are: (2, 0), (4, 0), (0, 5), (0, 10), (2, 5), (4, 5), (2, 10), (4, 10)
(g) Find all elements (a, b) ∈ Z6 ⊕ Z15 of order 15.
Answer:
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Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
Orders of a ∈ Z6 must divide 6, hence |a| = 1, 2, 3, 6.
Orders of b ∈ Z15 must divide 15, hence |b| = 1, 3, 5, 15.
Possibilities: lcm(1, 15) = lcm(3, 15) = lcm(3, 5) = 15.
Elements a ∈ Z6 such that |a| = 3 are a = 2, 4.
Elements b ∈ Z15 such that |b| = 5 are b = 3, 6, 9, 12.
Elements b ∈ Z15 such that |b| = 15 are b = 1, 2, 4, 7, 8, 11, 13, 14.
Elements (a, b) ∈ Z6 ⊕ Z15 of order 15 are the pairs:
(a, b) with a = 0, 2, 4 ∈ Z6 and b = 1, 2, 4, 7, 8, 11, 13, 14 ∈ Z15 and also
(a, b) with a = 2, 4 ∈ Z6 and b = 3, 6, 9, 12 ∈ Z15
(h) Find all elements (a, b) ∈ Z6 ⊕ Z15 of order 30.
2. Consider external direct product: Z6 ⊕ Z12 .
(a) Find a subgroup of order 4 which is cyclic.
Answer:
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To find a cyclic subgroup of order 4, need to find an element of order 4.
Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
Orders of a ∈ Z6 must divide 6, hence |a| = 1, 2, 3, 6.
Orders of b ∈ Z12 must divide 12, hence |b| = 1, 2, 3, 4, 6, 12.
The only possibilities are: lcm(1, 4) = lcm(2, 4) = 4.
For example: |(0, 3)| = lcm(1, 4) = 4.
Therefore < (0, 3) > is a cyclic group of order 4.
• Another example: |(3, 3)| = lcm(2, 4) = 4.
Therefore < (3, 3) > is a cyclic group of order 4.
(b) Find a subgroup of order 4 which is not cyclic.
Answer:
• Subgroup of order 4, which is not cyclic has no elements of order 4. So all elements
must have order 1 or 2.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
• The only element of order 1 is (0, 0).
• Elements of order 2 are:
(0, 6) since |(0, 6)| = lcm(1, 2) = 2
(3, 0) since |(3, 0)| = lcm(2, 1) = 2
(3, 6) since |(3, 6)| = lcm(2, 2) = 2.
• Check that the set H := {(0, 0), (0, 6), (3, 0), (3, 6)} forms a subgroup of Z6 ⊕ Z12 .
– By definition all elements of H are in Z6 ⊕Z12 , therefore H is a subset of Z6 ⊕Z12 .
– Since H is a finite subset, it is enough to check that it is closed under operation.
This can be checked by forming the Cayley table and noticing that no new
elements are introduced:
(0,0)
(0,6)
(3,0)
(3,6)
(0,0)
(0,0)
(0,6)
(3,0)
(3,6)
(0,6)
(0,6)
(0,0)
(3,6)
(3,0)
(3,0)
(3,0)
(3,6)
(0,0)
(0,6)
(3,6)
(3,6)
(3,0)
(0,6)
(0,0)
3. Is Z6 ⊕ Z15 isomorphic to Z90 ?
Answer:
• Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
• Orders of a ∈ Z6 must divide 6, hence |a| = 1, 2, 3, 6.
• Orders of b ∈ Z15 must divide 15, hence |b| = 1, 3, 5, 15.
• The only possibilities for the orders |(a, b)| are:
lcm(1, 1) = 1, lcm(1, 3) = 3, lcm(1, 5) = 5, lcm(1, 15) = 15,
lcm(2, 1) = 2, lcm(2, 3) = 6, lcm(2, 5) = 10, lcm(2, 15) = 30,
lcm(3, 1) = 3, lcm(3, 3) = 3, lcm(3, 5) = 15, lcm(3, 15) = 15,
lcm(6, 1) = 6, lcm(6, 3) = 6, lcm(6, 5) = 30, lcm(6, 15) = 30.
• Since no elements in Z6 ⊕ Z15 have order 90 and Z90 has elements of order 90 (for
example 1 ∈ Z90 ) these two groups cannot be isomorphic. Isomorphic groups have the
same numbers of elements of the same order.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
4. Is Z2 ⊕ Z45 isomorphic to Z90 ?
Answer:
• Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
• |(1, 1)| = lcm(|1|, |1|) = lcm(2, 45) = 90
• Therefore < (1, 1) > is cyclic group of order 90.
• < (1, 1) > is a subgroup of Z2 ⊕ Z45 , which is also a group with 90 elements.
• Therefore < (1, 1) >= Z2 ⊕ Z45 .
• Therefore Z2 ⊕ Z45 is a cyclic group of order 90.
• Every cyclic group of order 90 is isomorphic to Z90 .
• Therefore Z2 ⊕ Z45 is isomorphic to Z90 .
5. Let G = Z30 ⊕ Z20 .
(a) Determine the number of elements of order 15.
Answer:
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Order of (a, b) is |(a, b)| = lcm(|a|, |b|).
lcm(|a|, |b|) = lcm(15, 1) = lcm(15, 5) = lcm(3, 5)
Elements a ∈ Z30 with |a| = 15 are: 2,4,8,14,16,22,26,28.
Elements a ∈ Z30 with |a| = 3 are: 10,20.
Elements b ∈ Z20 with |b| = 5 are: 4,8,12,16.
Elements of order 15:
(a, b) with a = 2, 4, 8, 14, 16, 22, 26, 28 ∈ Z30 and b = 0, 4, 8, 12, 16 ∈ Z20 and also
(a, b) with a = 10, 20 ∈ Z30 and b = 4, 8, 12, 16 ∈ Z20 .
• #{elements of order 15} = ϕ(15)ϕ(1) + ϕ(15)ϕ(5) + ϕ(3)ϕ(5) = 8 · 1 + 8 · 4 + 2 · 4 = 48
(b) Determine the number of cyclic subgroups of order 15.
Answer:
• #{elements of order 15} = 48 (from part (a))
• Each cyclic subgroup of order 15 has ϕ(15) = 8 distinct generators.
• Two distinct cyclic subgroups of order 15, have distinct generators, i.e. elements of
order 15: the reason is that if they had a generator in common they would be the
same sungroup.
#{elements of order 15}
48
• #{cyclic subgroups of order 15}= {#generators
= ϕ(15)
= 48/8
of a cyclic group of order 15}
• #{cyclic subgroups of order 15} = 6
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
6. Let G = G1 ⊕ G2 . Let H1 be a subgroup of G1 . Prove that H1 ⊕ G2 is a subgroup of G.
Proof:
• H1 ⊕ G2 = {(h, g) | h ∈ H1 ,
g ∈ G2 }
• Claim 1: H1 ⊕ G2 6= ∅.
Proof of Claim 1: Let e1 ∈ G1 and e2 ∈ G2 be the identities of G1 and G2 .
The identity e1 of G1 is in H1 since H1 is a subgroup of G1 .
Therefore: (e1 , e2 ) ∈ H1 ⊕ G2 .
Therefore H1 ⊕ G2 6= ∅.
• Claim 2: H1 ⊕ G2 ⊂ G1 ⊕ G2 .
Proof of Claim 2: Let x ∈ H1 ⊕ G2 .
Then x = (h, g) for some elements h ∈ H1 and g ∈ G1 . Then h ∈ G1 since H1 ⊂ G1 .
Therefore x = (h, g) ∈ G1 ⊕ G2 .
• Claim 3: H1 ⊕ G2 is closed under group operation.
Proof of Claim 3: Let x, y ∈ H1 ⊕ G2 .
Then x = (h, g) and y = (k, d) with h, k ∈ H1 and g, d ∈ G2 .
Then xy = (h, g)(k, d) = (hk, gd) ∈ H1 ⊕ G2 .
Reason: hk ∈ H1 since H1 is a subgroup, hence closed under operation, and
gd ∈ G2 since G2 is a group, hence closed under operation.
Therefore xy ∈ H1 ⊕ G2 , hence H1 ⊕ G2 is closed under operation.
• Claim 4: H1 ⊕ G2 is closed under inverses.
Proof of Claim 4: Let x ∈ H1 ⊕ G2 .
Then x = (h, g) with h ∈ H1 and g ∈ G2 .
Then x−1 = (h−1 , g −1 ) ∈ H1 ⊕ G2 , since both H1 , G2 are closed under inverses (subgroup
and group properties).
Therefore x−1 ∈ H1 ⊕ G2 , hence H1 ⊕ G2 is closed under inverses.
• By Two step subgroup theorem (Nonempty subset, closed under operation and inverses
is a subgroup), it follows that H1 ⊕ G2 is a subgroup of G1 ⊕ G2 .
7. Let G = G1 ⊕ G2 . Prove that H = {eG1 } ⊕ G2 is a normal subgroup of G. Proof:
• You may use previous problem to prove that H = {eG1 } ⊕ G2 is a subgroup of G =
G1 ⊕ G2 .
• To show that it is a normal subgroup, it is enough to show:
xHx−1 ⊂ H for all x ∈ G, which is the same as:
x({eG1 } ⊕ G2 )x−1 ⊂ {eG1 } ⊕ G2 for all x ∈ G = G1 ⊕ G2 .
Proof:
– Let z ∈ xHx−1 = x({eG1 } ⊕ G2 )x−1 .
– Then z = x(eG1 , g)x−1 for some x = (a1 , a2 ) where a1 ∈ G1 and a2 ∈ G2 .
−1
– Then z = (a1 , a2 )(eG1 , g)(a1 , a2 )−1 = (a1 , a2 )(eG1 , g)(a−1
1 , a2 ), using that inverse in
the product can be found by finding inverse of each coordinate.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
−1
−1
−1
– Then z = (a1 eG1 a−1
1 , a2 ga2 ) = (eG1 , a2 ga2 ) ∈ {eG1 } ⊕ G2 . Reason: a2 , g, a2 ∈ G2
and since G2 is group it is closed under inverses and multiplications, so a2 ga−1
2 ) ∈ G2 .
– Therefore z ∈ H = {eG1 } ⊕ G2 , and hence
x({eG1 } ⊕ G2 )x−1 ⊂ {eG1 } ⊕ G2 , which is the same as xHx−1 ⊂ H.
• By the theorem stated in class xHx−1 ⊂ H for all x ∈ G is one of the equivalent
conditions for a subgroup H in G to be normal subgroup in G.
8. Let G = Z12 be the group under addition (mod12).
Let H =< 9 > be the subgroup generated by 9.
(a) What are the elements of H?
• H = {0, 9, 6, 3} = {0, 3, 6, 9}
(b) What is the size of H?
• |H| = |{0, 9, 6, 3}| = 4
(c) What is the size of each of the left cosets of H in G.
• |a + H| = |H| = 4
(Notice additive notation!)
(d) Find all left cosets of H in G.
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0+H
1+H
2+H
3+H
4+H
5+H
= {0, 3, 6, 9}
= {1, 4, 7, 10}
= {2, 5, 8, 11}
= {3, 6, 9, 0} = 0 + H = 6 + H = 9 + H
= {4, 7, 10, 1} = 1 + H = 7 + H = 10 + H
= {5, 8, 11, 2} = 2 + H = 8 + H = 11 + H
(e) How many left cosets of H in G are there?
• # of (distinct) left cosets of H in G = [G : H] = |G|/|H| = 12/4 = 3
• Also notice that we computed above 3 distinct left cosets.
(f) What is the size of each of the right cosets of H in G.
• |H + a| = |H| = 4
(Notice additive notation!)
(g) Find all right cosets of H in G.
• H + 0 = {0, 3, 6, 9} = H + 3 = H + 6 = H + 9
• H + 1 = {1, 4, 7, 10} = H + 4 = H + 7 = H + 10
• H + 2 = {2, 5, 8, 11} = H + 5 = H + 8 = H + 11
(h) How many right cosets of H in G are there?
• # of (distinct) right cosets of H in G = [G : H] = |G|/|H| = 12/4 = 3
• Also notice that we computed above 3 distinct right cosets.
(i) Are the left cosets of H in G the same as the right cosets of of H in G ?
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
• YES.
• 0 + H = H + 0, 1 + H = H + 1, 2 + H = H + 2, and also
a + H = H + a for all a ∈ G = Z12 .
(j) Prove that H is a normal subgroup of G.
• By definition a subgroup H in G is normal if all left cosets are the as right cosets,
i.e. aH = Ha for all a ∈ G, or in additive notation a + H = H + a for all a ∈ G.
• From the above computations it follows that H =< 9 > is normal subgroup in
G = Z12 .
(k) What are the elements of the quotient group G/H?
• Elements of the quotient group G/H are the left (or right) cosets.
• G/H = {0 + H, 1 + H, 2 + H}
(l) Write the Cayley table for G/H.
0+H
0+H 0+H
1+H 1+H
2+H 2+H
1+H
1+H
2+H
0+H
7
2+H
2+H
0+H
1+H
S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
9. Let G = S4 be the permutation group on four elements {1, 2, 3, 4}.
Let H =< (1342) > be the subgroup generated by the permutation α = (1342).
(a) What are the elements of H?
• H = {(1342), (14)(32), (1243), (1)}
(b) What is the size of H?
• |H| = |{(1342), (14)(32), (1243), (1)}| = 4
(c) What is the size of each of the left cosets of H in G.
• |aH| = |H| = 4
(d) Find all left cosets of H in G.
• (1)H = {(1342), (14)(32), (1243), (1)} = (1342)H = (14)(32)H = (1243) = H
• (12)H = {(12)(1342), (12)(14)(32), (12)(1243), (12)(1)} = {(134), (1423), (243), (12)}
• etc.
(e) How many distinct left cosets of H in G are there?
• # of (distinct) left cosets of H in G = [G : H] = |G|/|H| = 24/4 = 6
(f) What is the size of each of the right cosets of H in G.
• |Ha| = |H| = 4
(g) Find all right cosets of H in G.
• H(1) = {(1342), (14)(32), (1243), (1)} = H(1342) = H(14)(32) = H(1243) = H
• H(12) = {(1342)(12), (14)(32)(12), (1243)(12), (1)(12)} = {(234), (1324), (143), (12)}
• etc.
(h) How many right cosets of H in G are there?
(i) Are the left cosets of H in G the same as the right cosets of of H in G ?
• No! (12)H 6= H(12) from the above computations.
(j) Prove that H is not a normal subgroup of G.
• By definition a subgroup H in G is normal if all left cosets are the as right cosets,
i.e. aH = Ha for all a ∈ G.
• From the above computations it follows that H =< (1342) > is not normal subgroup
in G = S4 .
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
10. Let G = U (12) be the group of invertible integers (mod12).
Let H =< (5) > be the subgroup generated by the element (5).
(a) What are the elements of H?
(b) What is the size of H?
(c) What is the size of each of the left cosets of H in G.
(d) Find all left cosets of H in G.
(e) How many left cosets of H in G are there?
(f) What is the size of each of the right cosets of H in G.
(g) Find all right cosets of H in G.
(h) How many right cosets of H in G are there?
(i) Are the left cosets of H in G the same as the right cosets of of H in G ?
(j) Prove that H is a normal subgroup of G.
(k) What are the elements of the quotient group G/H?
(l) Write the Cayley table for G/H.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
11. Let G = A4 be the group of even permutations on four elements {1, 2, 3, 4}.
Let H =< (134) > be the subgroup generated by the permutation α = (134).
(a) What are the elements of H?
(b) What is the size of H?
(c) What is the size of each of the left cosets of H in G.
(d) Find all left cosets of H in G.
(e) How many left cosets of H in G are there?
(f) What is the size of each of the right cosets of H in G.
(g) Find all right cosets of H in G.
(h) How many right cosets of H in G are there?
(i) Are the left cosets of H in G the same as the right cosets of of H in G ?
(j) Prove that H is not a normal subgroup of G.
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
12. Prove that a group of order 25 has an element of order 5.
Proof:
• Let G be a group with |G| = 25. The order of each element of a group G must divide
the order of G (Theorem from the class). Therefore the only possible orders of elements
of G are: 1, 5, 25.
• The identity e is the only element in G with |e| = 1.
• Since G has more then one element, there must be another element a ∈ G and a 6= e.
• Since a 6= e, we have that |a| =
6 1.
• Therefore |a| = 5 or |a| = 25.
• If |a| = 5 we are done, since we wanted to show that there is an element in G of order 5.
• If |a| = 25, let b = a5 . Then |b| = |a5 | = 25/5 = 5. (Remember the rule for the order of
elements in cyclic groups: |ak | = n/k in cyclic group < a >, where |a| = n if k|n.)
• Therefore G has an element of order 5.
13. Let G be a group of order pq, where p and q are primes. Prove that any proper subgroup of
G is cyclic.
Proof:
• Let G be a group with |G| = pq, where p, q are primes.
• Order of any subgroup of G must divide order of the group |G| = pq.
• The only possible orders of subgroups of G are: 1, p, q, pq.
• Let H be a proper subgroup of G. Then |H| =
6 pq. Then |H| = 1, p, q.
• If |H| = 1 then H = {e} =< e >, hence H is cyclic.
• |H| = p or |H| = q then H is cyclic since every group of prime order is cyclic.
(As stated in class.)
14. Give an example of a group G which has a non-cyclic proper subgroup.
Some examples: In each example you have to explain why is H a subgroup and why it is not
cyclic.
• H = {(0, 0), (1, 0), (0, 3), (1, 3)} < Z2 ⊕ Z6
Check that all elements have order 2, hence H is not cyclic.
• H = 5Z20 ⊕ 4Z8 < Z20 ⊕ Z8
• H = 5Z200 ⊕ 4Z8 < Z200 ⊕ Z8
• H = {(1), (12)(34), (13)(24), (14)(23)} < S4
• H = {(1), (12)(34), (13)(24), (14)(23)} < A4
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S11MTH 3175 Group Theory (Prof.Todorov) Quiz 4 Practice (Some Solutions) Name:
15. Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, prove that G is
cyclic.
Proof:
• Elements of G have orders: 1,3,5,15.
• |e| = 1 Identity is the only element of order 1.
• Let H be a subgroup of order 3.
Since 3 is prime, H is cyclic, hence H =< a > with |a| = 3.
Elements of H are H = {a, a2 , e}.
• Let K be a subgroup of order 5.
Since 5 is prime, K is cyclic, hence K =< b > with |b| = 5.
Elements of K are K = {b, b2 , b3 , b4 , e}.
• H ∪ K = {e, a, a2 , b, b2 , b3 , b4 } ⊂ G
• Since |G| = 15 there must be some other elements in G.
• Let c ∈ G\(H ∪ K), i.e. c ∈ G but c ∈
/ H ∪ K.
– If |c| = 1 then c = e ∈ H ∪ K.
Therefore |c| =
6 1.
– If |c| = 3 then | < c > | = 3 and < c >6= H (since
This is contradiction to the assumption that H
Therefore |c| =
6 3.
– If |c| = 5 then | < c > | = 5 and < c >6= K (since
This is contradiction to the assumption that K
Therefore |c| =
6 5.
c∈
/ H).
is the only subgroup of order 3.
c∈
/ K).
is the only subgroup of order 5.
• Since |c| =
6 1, 3, 5, it follows that |c| = 15.
• | < c > | = 15, < c >⊂ G and |G| = 15 implies < c >= G.
• Therefore G is cyclic group.
q.e.d.
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