Solutions to Test #2 (Nov 21, 2011)
(1) For brevity, drop the subscripts. Let u be the transposition (12).
One of the cosets of A in S is the coset A itself. Another coset is
uA, since u is not in A. We must show that there are no other cosets.
Let tA be any coset of A in S. If t is even, then tA = A.
If t is odd, then tA=uA because ut is in A (since
the product of two odd permutations
is even). Thus there are no other cosets besides A and sA.
ALTERNATIVE PROOF: Produce a bijection between the odd perms and the even
perms, thus showing that A as well as every other coset has n!/2 elements.
Consequently, there must exist exactly two different cosets.
(2) Conjugation of F by either (R^k)F or (R^k) yields (R^2k)F, so
the conjugates of F are the four elements
(R^2m)F for m = 0, 1, 2, 3.
(3) Assume for the purpose of contradiction that G/Z is cyclic,
where Z denotes the center of G. Then there exists g in G such that
all cosets of Z in G have the form (g^j)Z for some integer j.
(Here gZ is an element of the quotient group G/Z which generates G/Z.)
Since G is a union of its cosets, every element of G has the form (g^j)z
for some integer j and some z in Z. It is easy to see that any two
elements of this form commute. Hence G is abelian, a contradiction.
(4) f maps G ONTO G because for any g in G, f(g^16) = g^32 = g.
It follows automatically then that f must be 1-1 as well, but alternatively,
one could prove f is 1-1 as follows. Assume f(a)=f(b). Then a^2 = b^2.
Since G is abelian, (b'a)^2 = 1, where b' denotes the inverse of b.
Raise both sides to the 16th power to see that b'a = 1, so a=b; thus f is 1-1.
To show that f is operation preserving, we can show f(ab)=f(a)f(b) as
follows: (ab)^2 = abab = a^2*b^2 because G is abelian.
(5a)
Let a be any element of A. Since C is contained in A,
aCa' is contained in aAa' which in turn is contained in A.
Also, aCa' is contained in aBa' which (by normality) is contained in B.
Thus aCa' is contained in C, the intersection of A and B.
Therefore C is normal in A.
(5b) Let a, x be in A, and let b, y be in B.
To show f is well-defined, we must show that if ab=xy, then aC=xC.
First note that ab=xy implies x'a = yb'.
The left side is in A and the right side
is in B, so x'a is in both A and B, so x'a is in C. This proves that
aC = xC (by the criterion that tells when two cosets are equal).
(6) EC: For brevity, let G denote the symmetric group S_8.
The number of conjugates of s=(1234)(5678) in G is 8!/(4*4*2) = 8!/32.
Thus the index |G:C(s)| = 8!/32, where C(s) denotes the centralizer of s
in G. Thus 8!/32 = |G|/|C(s)| =8!/|C(s)|, so |C(s)|=32. In other words,
exactly 32 elements of G commute with s.