ASSIGNMENT 3
Due Date: Oct 27,2010
i) Prove that a group of order 150 cannot be simple.
ii) Show that a finite group G can never be the union of the conjugates of a proper subgroup H ⊂ G.
iiii) A subgroup M of a group G is called a maximal subgroup if M 6= G and the only
subgroups of G which contain M are M and G. Analogously, a subgroup N of G is a
minimal normal subgroup if N is normal in G and N contains no other non-trivial normal
subgroup of G.
a) Show that if G =< x > is a cyclic group of order n ≥ 1, then a subgroup H is maximal
if and only if H =< xp > for some prime p dividing n.
b) Prove that the field of Q of rational numbers has no maximal subgroup.
c) Prove that every minimal normal subgroup of a finite solvable group is an abelian
group with the property that for any of its elements x, xp = 1, and also show that every
maximal subgroup of a finite solvable group has prime power index.
iv) Prove that if p is a prime and G is a group with | G |= p3 , then G is solvable.
v) a) Prove that the set F (S) of reduced words on S with concatenation as the binary
operation forms a group, and that F (S) is a free group on S. Also show that F (S) is
unique up to a unique isomorphism which is the identity map on S.
Note the following terminology: Let S be a subset of a group G such that G =< S >. A
presentation for G is a pair (S, R), where R is a set of words such that the smallest normal
subgroup containing < R > in the free group F (S) equals the kernel of the homomorphism
π : F (S) → G. The group G is finitely generated if the set S is finite, and G is finitely
presented if there is a presentation < S, R > with both S and R finite sets.
b) Prove that if | S |> 1, then F (S) is not abelian.
Finally, here is the fundamental theorem for finitely generated abelian groups:
Theorem: Let G be a finitely generated abelian group. Then
G ' Zr × Gt
where Gt is a finite abelian group. The integer r is called the free rank of G and the
cardinality n of Gt is the torsion cardinality of G.
k
Let n be the cardinality of Gt and suppose n = pα1 1 pα2 2 · · · pαk . Then there is an isomorphism
(1)
Gt ' A1 × A2 × · · · Ak .
where cardinality of Ai equals pαi i .
For simplicity, fix an i and write Ai = A, pi = p and αi = α. Let | A |= pα ; then there is
1
an isomorphism
(2)
A ' Z/pβ1 × Z/pβ2 × · · · × Z/pβt ,
with β1 ≥ β2 ≥ · · · ≥ βt ≥ 1, and β1 + · · · + βt = α. Of course t and βj0 s depend on i. The
integers pβi are called the elementary divisors of G.
vi) Prove that the decomposition in (1) and (2) is unique.
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