Solutions to 100A Final, Dec 6, 2011
(1) A_4 consists of the 12 even permutations (12)(34), (13)(24), (14)(23),
(123), (321), (124), (421), (134), (431), (234), (432), and the Identity.
D_4 consists of the 4 rotations R, R^2, R^3, R^4 and the 4 flips
FR, FR^2, FR^3, FR^4.
S_3 consists of the 6 permutations (12), (13), (23), (123), (321),
and the Identity.
Z(D_4) consists of R^2 and the Identity R^4.
Z(S_3) consists only of the identity.
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(2) True. Let g be a generator of C_200. That means g has order 200.
All generators have the form g^k, where k is relatively prime to 200.
It remains to show that phi(200)=80. This is true because
phi(200)=phi(25)*phi(8)=20*4=80.
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(3) Consider the 3 disjoint cosets H, 3H, and 9H.
The first coset consists of all elements of the form 3^k
where k = 0 mod 3; the second consists of all elements of the form 3^k
where k = 1 mod 3; and the third consists of all elements of the form
3^k where k = 2 mod 3. This exhausts all possible integers k,
and so G is the union of these three cosets. Since by definition
the index |G:H| equals the number of cosets, this index is 3.
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(4) There are (6*5)/2 = 15 transpositions.
There are (6*5*4*3)/(2*2*2!)=45 products of two transpositions.
There are (6*5*4*3*2*1)/(2*2*2*3!)=15 products of three transpositions.
The total is 15 + 45 + 15 = 75.
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(5A) For brevity let g' denote the inverse of g.
Note that (cd)' = d'c' = c'cd'c' = c'(cd'c') is in CD because
(cd'c') is in D by normality of D.
(5B) Suppose c1*d1 = c*d. We must show that f(c1*d1) = f(c*d),
otherwise f would not be well-defined. By definition of f,
we must show that cE = (c1)E, i.e., we must show x:=c'*c1 is in E.
From the first sentence, we know that x = d*(d1)', hence x is in both
C and D. This shows that x is in E, by definition of E.
(5C) True. The kernel of f is the set of elements cd for which
f(cd) = the identity coset E. Thus the kernel consists of all elements
cd for which cE=E, i.e., all elements cd for which c is in E.
The set of elements cd with c in E is precisely the set D, so the
kernel of f is D.
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(6) The other eight noncyclic abelian groups of order 144 are
C3 x C3 x C2 x C2 x C2 x C2
C3 x C3 x C2 x C2 x C4
C3 x C3 x C2 x C8
C3 x C3 x C4 x C4
C9 x C2 x C2 x C2 x C2
C9 x C2 x C2 x C4
C9 x C2 x C8
C9 x C4 x C4
Note that C9 x C16 is not included here, because that group is cyclic.
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(7) See pages 335 and 338 of the text.
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(8) Let A be the Sylow 3-subgroup and let B be the Sylow 11-subgroup.
Note A has 9 elements and B has 11 elements. The number of conjugates
k(A) of A is a divisor of 99 which is congruent to 1 mod 3, so k(A)=1.
The number of conjugates k(B) of B is a divisor of 99 which is congruent
to 1 mod 11, so k(B)=1. Thus both A and B are normal in G (since having
only one conjugate is equivalent to normality).
Since A and B have relatively prime orders, their intersection is {1}.
Hence AB is isomorphic to the direct product A x B, which has 99 elements.
Thus G=AB so G is isomorphic to A x B. All groups of
order p or p^2 are abelian, so both A and B are abelian. Thus
A x B is abelian, so G is abelian.
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