Section 5 5.1. In each case, determine whether or not H is a ...

Section 5
5.1. In each case, determine whether or not H is a subgroup of G.
a) G = (R, +); H = Q
b) G = (Q, +); H = Z
c) G = (Z, +); H = Z +
d) G = (Q − {0}, ·); H = Q +
e) G = (Z 8 , ⊕); H = {0, 2, 4}
f) G = the set of 2-tuples of real numbers (a, b) under addition of 2-tuples; H = the subset
consisting of all 2-tuples such that b = −a
g) G = Q 8 ; H = {I, J, K}
h) G = (P(X), △); H = {∅, A, B, A△B}, where A, B are two elements of G
i) G = (P(X), △); H = P, where Y ⊆ X.
a), b), d), f), h) and i) are subgroups. c) is not closed under inverses, and e) and g) are not
closed under the operation: 2 ⊕ 4 = 6 and JK = L.
5.4. a) How many subgroups does (Z 18 , ⊕) have What are they
The element 0 and the proper divisors of 18 generate the subgroups: 〈0〉, 〈1〉, 〈2〉, 〈3〉, 〈6〉
and 〈9〉, for a total of 6.
5.7. Let G = 〈x〉 be a cyclic group of order n. Show that x m is a generator of G if and only if
(m, n) = 1. Thus the number of generators of a cyclic group of order n is the number of
integers m in the set {0, 1, . . . , n − 1} such that (m, n) = 1. This number is denoted by φ(n);
φ is called Euler’s function and plays a prominent role in number theory.
x m is a generator of 〈x〉 if and only if 〈x m 〉 = 〈x〉, i.e., if and only if o(x m ) = n, i.e., if and
only if n/ gcd(m, n) = n, i.e., if and only if gcd(m, n) = 1.
5.11. Let G be an abelian group, and let n be a positive integer. Let H be the subset of G consisting
of all x ∈ G such that x n = e. Show that H is a subgroup of G.
Clearly e n = e, so e ∈ H. If x ∈ H, i.e., x n = e, then (x −1 ) n = (x n ) −1 = e −1 = e, so x −1 ∈ H.
If x, y ∈ H, i.e., x n = e and y n = e, then because G is abelian, (xy) n = x n y n = ee = e, so
xy ∈ H. Therefore, H is a subgroup of G.
5.12. Find the center of a) V ; b) Q 8 .
The center of V is V itself, because V is abelian. The center of Q 8 is {I, −I}, because J, K, L
and their negatives do not commute with each other.
5.13. Let H be the group introduced in Example 7 on p. 46. Find Z(H).
[ {( )
}]
a b
H =
: a, b, d ∈ R , ad ≠ 0
0 d
Suppose
( x y
0 z
)
∈ Z(H). Then for all
( x y
0 z
) ( a b
0 d
( a b
0 d
)
=
1
)
∈ H, we have
( a b
0 d
) ( x y
0 z
)

i.e., ( xa xb + yd
0 zd
)
=
( ax ay + bz
0 dz
so xb + yd = ay + bz for every value of a, b, d (with a, d nonzero). If we take a = b = d = 1,
we see we must have x = z, and then ( taking ) a = 1, b = 0 and d = 2, we see that we must
x 0
have y = 0. It is easy to check that
commutes with every element of H, no matter
0 x
what (nonzero real number) x is, so
{( ) }
x 0
Z(H) =
: x ≠ 0 .
0 x
5.17. Suppose H is a nonempty finite subset of a group G and H is closed under inverses. Must H
be a subgroup of G Either prove that it must, or give a counterexample.
A counterexample is {1, −1} ⊆ Z.
5.18. a) Show that it is impossible for a group G to be the union of two proper subgroups.
Assume, by way of contradiction, that G = A ∪ B where A, B are proper subgroups. Pick
x ∈ G − A and y ∈ G − B; then we must have x ∈ B and y ∈ A. Then we claim that
xy is in neither A nor B, contradicting the assumption. If xy ∈ A, then x = (xy)y −1 ∈ A,
contradicting our choice of A; and if xy ∈ B, then y = x −1 (xy) ∈ B, contradicting our choice
of y. So xy ∈ G − (A ∪ B), the desired contradiction.
b) Give an example of a group that is the union of three proper subgroups.
V is the union of its three proper subgroups 〈a〉, 〈b〉 and 〈c〉.
5.21. Let G = 〈x〉 be a cyclic group of order n. Find a condition on the integers r and s that is
equivalent to 〈x r 〉 ⊆ 〈x s 〉.
We have: 〈x r 〉 ⊆ 〈x s 〉 if and only if x r ∈ 〈x s 〉, which is true if and only if r = st + nu for some
integers t, u, which is true if and only if gcd(s, n)|r.
5.22. Let G be a group. Prove that Z(G) is a subgroup of G
The identity of G commutes with every element of G, so it is in Z(G). If x, y ∈ Z(G), i.e., for
every g in G we have xg = gx and yg = gy, then for every g in G we have (xy)g = x(yg) =
x(gy) = (xg)y = (gx)y = g(xy), so xy ∈ Z(G). Finally, if x ∈ Z(G), i.e., xg = gx for every
g in G, then also xgx −1 = g and so gx −1 = x −1 g for every g in G, so also x −1 ∈ Z(G).
Therefore, Z(G) is a subgroup of G.
5.26. Let H be a subgroup of the group G and let N(H) = {a ∈ G | aHa −1 = H}. (See Exercise 5.25
for the definition of aHa −1 .) Prove that N(H) is a subgroup of G.
We have eHe −1 = H, so e ∈ G. If aHa −1 = H, then aH = Ha and so H = a −1 Ha =
a −1 H(a −1 ) −1 , so N(H) is closed under inverses. If aHa −1 = H and bHb −1 = H, then
(ab)H(ab) −1 = abHb −1 a −1 = aHa −1 = H, so N(H) is closed under the operation. Therefore,
N(H) is a subgroup of G.
)
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