- Let
,
a group.
Suppose
.
Then show that
.
- Let
be a group and
,
.
Prove
if and only if
implies
.
(1 step subgroup test)
- Let
be a group and let
.
Let
.
Prove:
. This
subgroup is called the centralizer of
in
.
- Suppose
is a group which has only one element
such that
.
Prove that
, for all
.
- (Finite Subgroup Test) Let
be a nonempty finite subset
of a group
such that
implies
. Then show that
is a subgroup of
.
- Show that the intersection of any collection of
subgroups of a group
is a subgroup.
- Let
be a group. Referring to exercises 3 and 6,
the subgroup
is called the
center of
. Describe in words what
is, i.e.,
without using intersections.
- Show that if
is a finite group, its multiplication
table is a Latin square, i.e., each element of
the group appears once and only once in each row and
in each column of the table.
David Joyner
2007-08-06