Math 302
Modern Algebra
Spring 2009
Homework Solutions
7.5
CONGRUENCE AND LAGRANGE’S THEOREM
1.
Let
K
be a subgroup of a group
G
. Prove: For all
a
∈
G
,
Ka
=
K
if and only if
a
∈
K
.
Proof:
Suppose
a
is an arbitrary element of
G
. We need to prove
Ka
=
K
if and only if
a
∈
K
.
Forward Direction.
Suppose
Ka
=
K.
We need to show
a
∈
K.
Note
Ka
=
{
ka

k
∈
K
}
. Since
K
is a subgroup, then
e
∈
K
. It follows that
a
=
ea
∈
Ka
. Since
Ka
=
K
, then
a
∈
Ka
implies
a
∈
K
, as required.
Backward Direction.
Suppose
a
∈
K.
We need to show
Ka
=
K.
To prove this, we need to show
(
∀
g
∈
G
) [(
g
∈
Ka
)
↔
(
g
∈
K
)]
.
Suppose
g
is an arbitrary element of
G
. First suppose
g
∈
Ka.
We need to show
g
∈
K.
Since
g
∈
Ka
, then
g
=
ka
for some
k
∈
K
.
Since
K
is a subgroup and
a
∈
K
and
k
∈
K
, then
g
=
ka
∈
K
, as required.
Now suppose
g
∈
K.
We need to show
g
∈
Ka.
To show this, we need to show that
g
=
ka
for some
k
∈
K
.
Find a
k
.
The element
k
must satisfy
g
=
ka.
Solving this equation for
k
, we obtain
k
=
ga
−
1
.
Choose
k
=
ga
−
1
.
Verify
k
satisfies:
(i) Domain:
k
∈
K
(ii) Propositional Function:
g
=
ka
Since
a
∈
K
, then
a
−
1
∈
K
since
K
is a subgroup.
Then
g
∈
K
and
a
−
1
∈
K
imply
k
=
ga
−
1
∈
K
,
as required.
Since
k
=
ga
−
1
, then
g
=
ka
, as
required.
Therefore
g
∈
Ka
, as required. Thus
Ka
=
K.
This proves the backward direction.
As an alternative proof for the backward direction, note that as in the proof of the forward
direction,
a
∈
Ka
. Since
a
∈
K
also, then
a
∈
K
∩
Ka
. Since
K
=
Ke
, then
K
is a right coset of
K
.
By Corollary 7.24, two right cosets of
K
are either disjoint or identical. Since
a
∈
K
∩
Ka
, then
K
and
Ka
are not disjoint, so must be identical. Thus
K
=
Ka
.
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 Spring '03
 Edwards
 Algebra, Congruence, Subgroup, Cyclic group, Coset, Lagrange’s Theorem, distinct right cosets

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