Homework #4
Solutions
Due: September 21, 2011
1. Two results concerning groups are (1): If 𝐻 and 𝐾 are subgroups of 𝐺, then 𝐻 ∩𝐾 is a
subgroup (Theorem 0.4.4), and (2): Every subgroup of a cyclic group is cyclic (Theorem
0.4.5). For each of the following subgroups 𝐻 and 𝐾 of a cyclic group 𝐺 = [𝑎], identify
𝐻 ∩ 𝐾 by giving a generator. That is, write 𝐻 ∩ 𝐾 = [𝑏] for some 𝑏 ∈ 𝐺.
(a) 𝐺 = ℤ, 𝐻 = 6ℤ, 𝐾 = 45ℤ.
▶ Solution. 𝐻 ∩ 𝐾 = {𝑚 ∈ ℤ : 6∣𝑚 and 45∣𝑚}. Thus 𝐻 ∩ 𝐾 is the set of all
common multiples of 6 and 45. Since 𝐻 ∩ 𝐾 is a subgroup of ℤ it follows, from
the Theorem proven in class, (or see the proof of Theorem 0.4.5, Page 25) that
𝐻 ∩ 𝐾 = 𝑏ℤ where 𝑏 is the smallest positive integer in 𝐻 ∩ 𝐾. That is, 𝑏 is the
least common multiple of 6 and 45, which is 90. Thus, 𝐻 ∩ 𝐾 = 90ℤ.
◀
(b) 𝐺 = [𝑎] with 𝑜(𝑎) = 20, 𝐻 = [𝑎14 ], 𝐾 = [𝑎15 ].
▶ Solution. Since 𝑜(𝑎14 ) = 20/ gcd(20, 14) = 10 (Theorem 0.3.4), it follows
that the order of the cyclic group 𝐻 is 10. According to Theorem 0.4.6, there is
a unique subgroup of 𝐺 of order 10, namely [𝑎20/10 ] = [𝑎2 ]. Thus, 𝐻 = [𝑎2 ]. In
other words, 𝑎2 is also a generator of 𝐻 along with 𝑎14 . (In fact, (𝑎14 )3 = 𝑎2 .)
Similarly, 𝑜(𝑎15 ) = 20/ gcd(15, 20) = 20/5 = 4. Thus, 𝐾 = [𝑎20/4 ] = [𝑎5 ]. Thus,
𝐻 = {𝑒, 𝑎2 , 𝑎4 , . . . , 𝑎18 } and 𝐾 = {𝑒, 𝑎5 , 𝑎10 , 𝑎15 } and hence, 𝐻∩𝐾 = {𝑒, 𝑎10 } =
[𝑎10 ].
◀
2. Find all of the subgroups of each of the following groups.
(a) ℤ18
▶ Solution. This is a direct application of Theorem 0.4.6. There is a unique
subgroup of ℤ18 for each positive divisor of 18 = ∣ℤ18 ∣. These divisors are 1, 2, 3,
6, 9, 18. For each divisor 𝑑, the subgroup of ℤ18 of order 𝑑 is (18/𝑑)ℤ18 . Hence,
the subgroups are {0}, 9ℤ18 , 6ℤ18 , 3ℤ18 , 2ℤ18 , and ℤ18 .
◀
(b) 𝐺 = [𝑎] where 𝑜(𝑎) = 28.
▶ Solution. The divisors of 28 are 1, 2, 4, 7, 14, 28. The corresponding subgroups are [𝑒], [𝑎14 ], [𝑎7 ], [𝑎4 ], [𝑎2 ], and [𝑎].
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(c) ℤ∗13 . Hint: Show first that ℤ∗13 is cyclic with generator 2.
▶ Solution. Since 13 is prime, ℤ∗13 = {1, 2, . . . , 12}. The powers of 2 mod 12
are 21 = 2, 22 = 4, 23 = 8, 24 = 3, 25 = 6, 26 = 12, 27 = 11, 28 = 9, 29 = 5,
210 = 10, 211 = 7, 2𝑙12 = 1. Thus, ℤ∗13 = [2] is cyclic of order 12, with generator 2.
Since the divisors of 12 are 1, 2, 3, 4, 6, 12, the subgroups of ℤ∗13 are [1] = [212 ],
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[26 ] = [12], [24 ] = [3], [23 ] = [8], [22 ] = [4], and [2] = ℤ∗13 .
3. (a) Find all of the left cosets of the subgroup 𝐻 = 4ℤ8 of the group 𝐺 = ℤ8 . Recall
that for additively written groups, such as ℤ8 , the cosets are written additively
in the form 𝑎 + 𝐻 = {𝑎 + ℎ : ℎ ∈ 𝐻}.
Math 4023
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Homework #4
Solutions
Due: September 21, 2011
▶ Solution. 0+𝐻 = {0, 4}, 1+𝐻 = {1, 5}, 2+𝐻 = {2, 6}, 3+𝐻 = {3, 7}. ◀
(b) Suppose that 𝐺 = [𝑎] with 𝑜(𝑎) = 12. Find all of the cosets of the subgroup
𝐻 = [𝑎4 ].
▶ Solution. 𝑒𝐻 = {𝑒, 𝑎4 , 𝑎8 }, 𝑎𝐻 = {𝑎, 𝑎5 , 𝑎9 }, 𝑎2 𝐻 = {𝑎2 , 𝑎6 , 𝑎10 }, 𝑎3 𝐻 =
{𝑎3 , 𝑎7 , 𝑎11 }.
◀
4. Suppose 𝐾 is a proper subgroup of 𝐻 and 𝐻 is a proper subgroup of 𝐺. (𝐻 is a proper
subgroup of 𝐺 means that 𝐻 is a subgroup but 𝐻 ∕= 𝐺.) If ∣𝐾∣ = 42 and ∣𝐺∣ = 420,
what are the possible orders of 𝐻.
▶ Solution. According to Lagrange’s Theorem (Theorem 0.4.9), ∣𝐾∣ ∣ ∣𝐻∣ and ∣𝐻∣ ∣ ∣𝐺∣.
Thus, ∣𝐻∣ = 42𝑚 and 42𝑚∣420. Hence, 420 = 42𝑚𝑡 so 𝑚𝑡 = 10 and 𝑚∣10. Thus,
𝑚 = 1, 2, 5, or 10. But 𝑚 ∕= 1 since 𝐻 ∕= 𝐾 (so it must have order > 42) and 𝑚 ∕= 10
since 𝐻 ∕= 𝐺 so ∣𝐻∣ ∕= 420. Thus, ∣𝐻∣ = 42𝑚 for 𝑚 = 2 or 𝑚 = 5. Thus, ∣𝐻∣ = 84 or
∣𝐻∣ = 210.
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5. Find the value of the Euler phi-function 𝜙(𝑛) for (a) 𝑛 = 97; (b) 𝑛 = 8800.
▶ Solution. (a) 𝜙(97) = 96 since 97 is prime.
(b) The prime factorization of 8800 is 8800 = 25 ⋅ 52 ⋅ 11. Hence,
𝜙(8800) =
=
=
=
=
𝜙(25 ⋅ 52 ⋅ 11)
𝜙(25 )𝜙(52 )𝜙(11)
(25 − 24 )(52 − 51 )(11 − 1)
16 ⋅ 20 ⋅ 10
3200.
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6. Use Euler’s Theorem (theorem 0.6.1) to find a number 𝑎 with 0 ≤ 𝑎 < 73 with
𝑎 ≡ 9794
(mod 73).
▶ Solution. Since 73 is prime, 𝜙(73) = 72. Dividing 794 by 72 gives 794 = 72 ⋅ 11 + 2.
Thus, Euler’s theorem gives 972 ≡ 1 (mod 73), so
9794 = 972⋅11+2 = (972 )11 92 ≡ 92 ≡ 81 ≡ 8 (mod 73).
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7. For each part, find the smallest positive 𝑥 that solves the given simultaneous congruences.
(a) 𝑥 ≡ 5 (mod 7) and 𝑥 ≡ 5 (mod 9)
Math 4023
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Homework #4
Solutions
Due: September 21, 2011
▶ Solution. There was a typo in this exercise. As stated, the obvious answer is
𝑥 = 5.
◀
(b) 𝑥 ≡ 5 (mod 5) and 𝑥 ≡ 2 (mod 12) and 𝑥 ≡ 8 (mod 13)
▶ Solution. Use the Chinese Remainder Theorem. The moduli 5, 12, 13 are
pairwise relatively prime. Let 𝑁1 = 12 ⋅ 13 = 156, 𝑁2 = 5 ⋅ 13 = 65, and
𝑁3 = 5 ⋅ 12 = 60. Use the Euclidean algorithm (or inspection) to write:
156 − 31 ⋅ 5 = 1
5 ⋅ 65 − 27 ⋅ 12 = 1
5 ⋅ 60 − 23 ⋅ 13 = 1.
Then, 𝑥 = 5 ⋅ 156 + 2 ⋅ 5 ⋅ 65 + 8 ⋅ 5 ⋅ 60 = 3830 is the unique solution of the 3
congruences modulo 𝑁 = 5 ⋅ 12 ⋅ 13 = 780. Reducing 3830 modulo 780 we get that
the smallest positive solution is 3830 − 4 ⋅ 780 = 710.
◀
Math 4023
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