2.2 Groups
1.
a. Not a group. Only 0 has an inverse so G4 fails.
c. Group. It is clearly closed and
proves associativity. The unity is −1, and the inverse of a is −a − 2. Note that G is also abelian.
e. Not a group. It is not closed: (1 2)(1 3) = (1 3 2) is not in G. Note that ε is a unity and each element is self inverse, so only G1 fails.
g. Group. The unity is 16; associativity from
. For inverses and closure —see the Cayley table:
i. Not a group. It is closed (by Theorem 3 §0.3), and associative, and
ε is the unity. However G4 fails. If
has
σn = 2
n for all
then
σ has no inverse because it is not onto.
3. a.First
ad =
c,
a2 =
d by the Corollary to Theorem 6. Next
ba ≠
b,
a,
d; and
ba =
c ⇒
b =
ac =
a(
ba) = (
ab)
a = 1
a =
a, a contradiction. So
ba = 1. Then
bd =
a,
bc =
d,
b2 =
c. Next,
ca =
b,
cd = 1,
c2 =
a,
cb =
d. Finally,
da =
c,
db =
a,
dc = 1,
d2 =
b.
5. A monoid is a group if each element is invertible. So check that every row and column contains exactly one 1.
7. The unity is
I3 and
shows that
G is closed. Since matrix multiplication in general is associative, it remains to show that each matrix in
G has an inverse in
G. But
as is easily verified.
8. a. Write σ = (1 2)(3 4), τ = (1 3)(2 4) and ϕ = (1 4)(2 3). Then σ2 = τ2 = ϕ2 = ε and στ = τσ = ϕ, σϕ = ϕσ = τ and ϕτ = τϕ = σ. Hence G is closed and every element is self inverse. Since permutation multiplication in general is associative, G is a group. Here x2 = ε for all four elements x of G.
9. It is easy to show that
and σ6 = ε. Hence G = {ε, σ, σ2, σ3, σ4, σ5} is closed by the exponent laws and σ−1 = σ5, (σ2)−1 = σ4, (σ3)−1 = σ3, (σ4)−1 = σ2 and (σ5)−1 = σ. Since permutation multiplication is associative, G is a group. Also, G is abelian because σkσl = σk+l = σlσk for all k, l. Finally, there are two elements τ satisfying τ2 = ε : τ = ε and τ = σ3; the three with τ3 = ε are τ = ε, τ = σ2 and τ = σ4.
10.
a. ab = ba2 gives aba2 = ba4 = b. Hence a2ba2 = ab, that is a2ba2 = ba2. Cancellation gives a2 = 1 . Then ab = ba2 = b, whence a = 1 by cancellation.
c. ab = ba2 gives aba4 = ba6 = b. Hence a2ba4 = ab = ba2, so a2ba2 = b by cancellation. Finally a3ba2 = ab = ba2 so a3 = 1. Hence b = aba4 = aba.
11. a. We claim that
b(
ab)
na = (
ba)
n+1 for all
n ≥ 0. It is clear if
n = 0. If it holds for some
n ≥ 0, then
Hence this holds for all n ≥ 0 by induction. Now suppose (ab)n = 1. Then (ba)n+1 = b(ab)na = b1a = ba. Cancelling ba gives (ba)n = 1.
13. α is onto because
g = (
g−1)
−1 =
α(
g−1) for all
g G. If
α(
g) =
α(
g1), then
, so
. This shows that
α is one-to-one.
15. Define σ : X → Xa by σ(x) = xa. This is clearly onto and σ(x) = σ(x1) implies xa = x1a, so x = x1 by cancellation. Hence σ is one-to-one.
17. If e2 = e, then ee = e1, so e = 1 by cancellation. Thus 1 is the only idempotent.
19. If
G is abelian, then
gh =
hg, so (
gh)
−1 = (
hg)
−1 =
g−1h−1 by Theorem 3. Conversely, given
x,
y G, we are assuming (
xy)
−1 =
x−1y−1. By Theorem 3, this is
y−1x−1 =
x−1y−1; that is any two inverses commute. But this means that
G is abelian because
every element
g of
G is an inverse [in fact
g = (
g−1)
−1].
21. If G is abelian, then (gh)2 = g(hg)h = g(gh)h = g2h2 for all g, h. Conversely, if (gh)2 = g2h2, then g(hg)h = g(gh)h. Thus hg = gh by cancellation (twice).
23. a. If g = g−1, then g2 = gg−1 = 1; if g2 = 1, then g−1 = g−11 = g−1g2 = g.
25. Let
a5 = 1 and
a−1ba =
bm. Then
Next
. This continues to give
and finally
. Hence
by cancellation.
27. In multiplicative notation,
a1 =
a,
a2 =
a ·
a,
a3 =
a·
a ·
a, . . .; in additive notation
a +
a = 2
a,
a+
a +
a = 3
a, . . .. In
,
, so
is generated by 1.
29.
b.We first establish left cancellation: If gx = gy in G, then x = y. In fact, let hg = e. Then gx = gy implies x = ex = hgx = hgy = ey = y. Thus hg = e = ee = hge, so g = ge by left cancellation. This shows that e is the unity. Finally, h(gh) = (hg)h = eh = h = he, so gh = e, again by left cancellation. Thus h is the inverse of g.
2. Choose
g G and let
ge =
g,
e G (by hypothesis). If
zg =
e,
z G, then
e =
zg =
zge =
ee =
e2. Now, given
h G, let
h =
ex. Then,
eh =
e2x =
ex =
h. Similarly,
h =
ye,
y G, implies
he =
h. Thus
e is the unity for
G. But now, given
h, we can find
c,
d such that
ch =
e =
hd. Then
c =
ce =
c(
hd) = (
ch)
d =
ed =
d, so
ch =
e =
hc. Thus
h has an inverse.