8.3.2.a
<(1234),(24)>={(1),(1234),(13)(24),(1432),(24),(12)(34),(13),(14)(32)}
<(1243),(23)>={(1),(1243),(14)(23),(1342),(23),(12)(43),(14),(13)(42)}
<(1234),(24)>={(1),(1324),(12)(34),(1423),(34),(13)(24),(12),(14)(23)}
8.3.5.b
Let n be the number of Sylow-5 subgroups of a group G of order 60.
Then n||G| and n=1+5k for some integer k.
The divisors of G are : 1,2,3,4,5,6,10,12,15,20,30.
Only 1 and 6 from this list satisfy the second condition.
So every group of order 60 has 1 or 6 Sylow-5 subgroups.
8.3.15.b
Let G be the intersection of H and K. Note that G is a subgroup of K.
Let n=|K:G| and let k1,k2,...,kn be right coset representatives of G in K. |K|=|G||K:G|,
therefore n=|K|/|G|.
Let hk be in HK. k is in K, so there is an integer i and g in G s.t. k=g*ki. So hk=h*g*ki.
h and g are in H so h1=hg is in H and hk=h1*k1.
We get that HK=union(H*ki) i from 1 to n. If H*Ki=H*kj then ki*(kj)^(-1) is in H. But ki's are in K
so ki*(kj)^(-1) is in G. This means G*ki=G*kj which contradicts to the definition of ki's.
So HK=disjoint union of H*ki.
Now |H*ki|=|H| and we get that |HK|=n|H|=|K||H|/|G|.
8.4.1.c
The conjugacy classes are:
{(123),(134),(421),(432)}
{(124),(234),(321),(431)}
{(12)(34),(13)(24),(14)(23)}
8.4.7.a
Let x be in A. Then for every a in A x-1ax is in A since A
is a group. So x-1Ax is a subset of A. If a and b are in A and
a!=b, then x-1ax!=x-1bx. This means
that |A|=|x-1Ax|. But x-1Ax is
contained in A, therefore x-1Ax=A. This means x is in N(A).
So we get A is a subset of N(A).
8.4.7.b
For every g in G, g is in N(A) iff g-1Ag=A iff
gg-1Ag=gA iff Ag=gA.
8.4.9
Let g,h be in f(C). Then there are a and b in C s.t. g=f(a) and h=f(b).
a and b are in C and C is a conjugacy class, so there is an x in G s.t.
a=x-1bx. f is an automorphism, so
g=f(a)=f(x)-1f(b)f(x)=f(x)-1hf(x).
This means g and h are conjugates.
Let g=f(a) be in f(C) and k be a conjugate of g. Then there is a t in G s.t.
k=t-1gt. Now f is an isomorphism, so there must be an r in
G s.t. f(r)=t. So we get k=f(r)-1f(a)f(r)=f(r-1ar).
r-1ar is a conjugate of a, and C is a conjugacy class. This implies that
r-1ar is in C, i.e. k is in f(C).
So every two elements in f(C) are conjugate to each other, and every conjugate of an element in f(C)
is again in f(C), therefore f(C) is a conjugacy class.
8.4.14.a
Let a be in N. N is normal, so for every g in G, g-1Ng=N, in particular
g-1ag is in N. So every conjugate of a is in N, i.e. C is a subset of N.
Let C be a subset of N. C={g in G | g is conjugate to a}. But a=e-1ae, so a
is conjugate to a, therefore a is in C. Since C is a subset of N, we get a is in N.