Algebra Prelim Solutions, Fall 2003
- Let G be a simple group of order
2256 = 16*3*47. The number of Sylow 47-subgroups is congruent to 1
mod 47 and divides 48. Since G is simple, the number of Sylow
47-subgroups is not 1 (otherwise G would have a normal Sylow
47-subgroup), consequently G has 48 Sylow 47-subgroups. Next the
number of Sylow 3-subgroups is congruent to 1 mod 3 and divides
16*47. This number cannot be 1 because G is simple. Nor can it be
4, because then G would be isomorphic to a subgroup of A4 which
only has order 12. We
conclude that G has at least 16 Sylow 3-subgroups. Finally G has
at least two Sylow 2-subgroups. Note that
two distinct Sylow 47-subgroups P, Q intersect trivially, because
| P /\ Q| divides 47 and is not 47. Similarly two distinct Sylow
3-subgroups intersect trivially. On the other hand we cannot assert
that two distinct Sylow 2-subgroups intersect trivially.
We now count elements. The number of elements of order 47 is
46*48; the number of elements of order 3 is at least 2*16, and
the number of elements of order a power of two is at least
17. We now have at least
17 + 2*16 + 46*48 = 2257 elements in G, which
is too many. This proves that there is no simple group of order
- Every element of G can be written in the form
ai, bi Z. Using
x7 = y3 = 1, we may assume that
ai, bi> 0 for all i. Now whenever
there is a subword yx, we can replace it with xy2, because
yxy-1 = x2, so we may push all the x's to the left and all
the y's to the right. This proves the result.
- Using (i) and
x7 = y3 = 1, we may write every element of G in
the form xiyj where
0 < i < 6 and
0 < j < 2. It
| G| < 7*3 = 21.
- This is because
q respects the relations. Specifically,
(1 3 4 5 6 7)7 = 1,
((2 3 5)(4 7 6))3 = 1,
((2 3 5)(4 7 6))(1 2 3 4 5 6 7)((2 3 5)(4 7 6))-1 = (1 3 5 7 2 4 6).
- We see from (iii) that
imq has elements of order 7 and 3.
Therefore 21 divides
| imq| and hence 21 divides | G|.
| G| < 21 from (ii), we deduce that | G| = 21 as required.
- We may assume that
a, b =/= 0.
Let d be the greatest common divisor of a, b. We claim that
a/d, b/d are coprime. If r divides a/d, b/d, then rd divides
a, b and hence rd divides d, say srd = d for some s R.
Therefore sr = 1, which means that r is a unit and the claim is
established. By hypothesis
(a/d )R + (b/d )R = xR for some x R. Then
aR + bR = xdR and the result follows.
- Note that
M @ Z/2Z as R-modules, so we
need to prove
Z/2Z Z/2Z @ Z/2Z. Define
q : Z/2Z Z/2Z - > Z/2Z by
q(x, y) = xy. It is easily verified that
q is an R-balanced map and
q induces an R-module map
q' : Z/2Z Z/2Z - > Z/2Z satisfying
q(x y) = xy. Now define
f : Z/2Z - > Z/2Z Z/2Z by
fx = x 1. Then
q'fx = q'(x 1) = x, so
q'f is the identity. Also
fq'(x y) = f(xy) = xy 1 = x 1. Therefore
the identity on a set of generators of
Z/2Z Z/2Z (in this case these generators
x y are the whole of
Z/2Z Z/2Z, but that won't always be the case) and we
fq' is also the identity. This proves that
Z/2Z Z/2Z @ Z/2Z as required.
- Note that N is finitely generated. This is because M is finitely
M M is finitely generated, which implies
(M M)/(N 0) @ N is finitely generated.
We now apply the structure theorem for finitely generated modules
over a PID. We may write
|| @ Ra (R/q1R)a1 ... (R/qnR)an
|| @ Rb (R/q1R)b1 ... (R/qnR)bn
n, a, b, ai, bi are non-negative integers and the qi are
nonassociate prime powers. Since
M M @ N N,
we see that
By the uniqueness part of the structure theorem, we deduce that
2a = 2b,
2a1 = 2b1, ...,
2an = 2bn. Therefore
a = b, a1 = b1, ..., an = bn and the result follows.
)2an @ R2b
- Note that f has degree 4. Since
ai satisfies f, we
[K(ai) : K] < 4. Therefore
[K(ai,aj) : K] < 4*4 = 16, in particular
K(ai,aj) =/= L as
- Assume that f is monic. Then if a K is the coefficient of
x3 of f, we have
a1 + a2 + a3 + a4 = - a.
Since L is the splitting field over K of the polynomial f and
we are in characteristic zero, L is a Galois extension of K. Let
G denote the Galois group of L over K. Then the subfields of
L containing K are in a one-to-one correspondence with the
subgroups of G, and G permutes the
ai faithfully. If
K(a1 + 2a2 + 3a3) =/= L, then
K(a1 + 2a2 + 3a3) corresponds to a nontrivial
subgroup of G, in particular
a1 + 2a2 + 3a3 is
fixed by some nontrivial permutation of the
ai. In other
words there is
1 =/= s S4 such that
a1 + 2a2 + 3a3 = as1 + 2as2 + 3as3.
this will yield a different linear relation from
a1 + a2 + a3 + a4 = - a from above.
(Clearly this relation is not 0 = 0. If we subtract the right hand
side from the left hand side, then the sum of the coefficients of the
ai is 0.) Therefore we may express the
ai in terms
of just two of the
ai; in other words,
L = K(ai,aj) for some i, j, which contradicts (i).
K[a1 + 2a2 + 3a3] = K(a1 + 2a2 + 3a3)
a1 + 2a2 + 3a3 is algebraic over K and the
- If the assertion is false, then
I(V1) + I(V2)
is contained in a maximal ideal M of
k[x1,..., xn]. By
Hilbert's Nullstellensatz, M is of the form
(x1 - a1,..., xn - an), where ai k for all i. Since
f (a1,..., an) = 0 for all f M, we see that
f (a1,..., an) = 0 for
f I(V1) and
(a1,..., an) V1 /\ V2 which is a
contradiction, and the result follows.