Download Math330 Fall 2008 7.34 Let g be a non

Math330
HW 8
Fall 2008
7.34 Let g be a non-identity element of G. Then |g| must be 12, 6, 4, 3, or 2. If the order of
g is 2 then we are done. If |g| = 12, 6, 4 then |g 6 | = 2, or |g 3 | = 2, or |g 2 | = 2. So the only
way for there not to be an element of order two in G is if all non-identity elements have
order 3. Let’s assume that this is true, and get a contradiction.
Assume that all 11 non-identity elements of G have order 3. Notice that elements of order
3 come in pairs, i.e. if |g| = 3 then |g 2 | = 3. But this is a contradiction, because we have
11 non-identity elements, which is an odd number. Thus G cannot have all non-identity
elements of order 3, and therefore must have an element of order 2.
8.16 We follow the examples 4 and 5 on page 155 in order to determine the number of
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elements of order 15 in z30 Z20 . By Theorem 8.1, we count the number of elements (a, b)
with the property that |(a, b)| = lcm(|a|, |b|) = 15.
Case 1: |a| = 15 and |b| = 1 or 5. There are 8 choices for a and 5 choices for b, so there
are 8 × 5 = 40 elements of order 15.
Case 2: |a| = 3 and |b| = 5. There are 2 choices for a and 4 choices for b, giving 8 additional
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elements of oder 15 in Z30 Z20 .
So there are a total of 48 elements of order 15. Since each cyclic subgroup of order 15
has 8 elements of order 15 and no two cyclic subgroups can have an element of order 15 in
common, there are 48/8 = 6 cyclic subgroups of order 15.
8.24. Notice that the question does not ask about cyclic subgroups, but subgroups in genL
eral. The cyclic subgroups of order 4 in Z4 Z4 are < (1, 0) >, < (1, 1) >, < (1, 2) >, <
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(1, 3) >, < (0, 1) >, < (2, 1) > . The other subgroup of order 4 in Z4 Z4 is {(0, 0), (2, 0), (0, 2), (2, 2)}.
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8.30. We want to determine if H is isomorphic to Z9 or Z3 Z3 . We can do this several
ways, one of which is to show directly that each non-identity element of H has order 3.
8.42. This is a special case of Theorem 8.2. We want to find an isomorphism from Z12 to
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Z4 Z3 . Let
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φ : Z12 → Z4
Z3
be the map
1 7→ (1mod4, 1mod3).
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This map is clearly well-defined. We can check that φ(x + y) = (x + ymod4, x + ymod3) =
(xmod4, xmod3)+(ymod4, ymod3). The map is injective since φ(x) = φ(y) ⇒ (xmod4, xmod3) =
(ymod4, ymod3) ⇒ xmod4 = ymod4 and xmod3 = ymod3 ⇒ x = y. The map φ is also
surjective.
9.8 a. By referring to the table 5.1 on page 105 we can check that α5 H = {α5 , α8 } =
6
Hα5 = {α5 , α6 }. Therefore by definition, H cannot be a normal subgroup of A4 .
9.8 b. The fact that α6 α9 H is not equal to α7 α11 H shows that coset multiplication is not
a binary operation (review the definition of binary operation on page 42). Thus the set of
left cosets of H cannot be a group.
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