7.9.1.a
(1 7 3)
7.9.2.b
(1 3 5 6 2 4 7)
7.9.3.d
(14)(27)(523)(34)(1472)= (1 2 3 4 5 6 7)=(1 5 7 3)(2 4)
(5 4 1 2 7 6 3)
7.9.5.a
(2468)=(28)(26)(24) odd
(246)(134)=(26)(24)(14)(13) even
(12)(123)(1234)=(12)(13)(12)(14)(13)(12) even
7.9.16
Note that disjoint cycles always commute.
Suppose Z(Sn)!={e}.
Let g be in Z(Sn). g!=e.
Then can be written as a product of disjoint cycles. Let
g=(a(1) ... a(k))...(b(1) ... b(m)).
Let g1:=(a(1) ... a(k)).
Case1: There is a cycle of length more than 2.
WLOG k is bigger than 2. Let h=(a(1) a(2)). Then we must have
gh=hg. Now h is disjoint from all the cycles in g except g1,
therefore, h commutes with all except g1. So h must commute
with g1 too. But an easy calculation shows that
gh=(a(1) a(3) a(4) ... a(k)) != (a(2) a(3) ... a(k)) = hg.
This is a contradiction.
Case2: All the cycles in g have length 2.
Case2a: There are more than 1 cycles in the cycle decomposition of g.
g=(a1 a2)(b1 b2)(c1 c2)...(d1 d2). Then
g(a1 b1)=(a1 b2 b1 a2)(c1 c2)...(d1 d2) !=
(a1 a2 b1 b2)(c1 c2)...(d1 d2) = (a1 b1)g. Again contradiction.
Case2b: g is a transposition. g=(a1 a2). Then
g(a2 a3)=(a1 a2 a3) != (a1 a3 a2) = (a2 a3)g.
In all cases we got a contradiction, therefore there is no g in Z(Sn)
s.t. g!=e.
7.9.33
For every a and b (a b)=(1 b)(1 a)(1 b).
We know that every permutation is a product of transpositions,
and as we showed here (1 2),(1 3), ... (1 n) generate
every transposition. So (1 2),(1 3), ... (1 n) generate all the
permutations in Sn.
8.1.1.a
element order
(0 0) 1
(0 1) 4
(0 2) 2
(0 3) 4
(1 0) 2
(1 1) 4
(1 2) 2
(1 3) 4
8.1.6.a
Z(12)={0,4,8}x{0,3,6,9}
8.1.9
Z(8) is not isomorphic to Z(4)xZ(2) since Z(8) has an element of order 8 but Z(4)xZ(2) doesn't.