Every group is normal in itself
This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., identity-true subgroup property)
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Contents
Statement
Statement with symbols
Suppose is a group. Consider
as a subgroup of itself --
is a normal subgroup of itself.
Related facts
Opposite facts
Similar facts
- Invariance implies identity-true
- Balanced implies identity-true
- Every group is characteristic in itself
- Trivial subgroup is normal
Proof
Proof using the conjugation definition of normality
Given: A group .
To prove: For any and
,
.
Proof: This is obvious from the fact that is closed under multiplication.
Proof using the kernel of homomorphism definition of normality
Given: A group .
To prove: There exists a homomorphism for some group
such that every element of
maps to the identity element.
Proof: Let be the trivial group and
be the map sending every element of
to the identity element of
. Clearly,
satisfies the conditions for being a homomorphism: for any
, both
and
equal the identity element of
. Moreover, every element of
is sent to the identity element of
under
.
Proof using the cosets definition of normality
Given: A group .
To prove: For every element ,
.
Proof: Note that
-
: Clearly,
. Also, for any
,
, so
. Thus,
.
-
: Clearly,
. Also, for any
,
, so
. Thus,
.
Combining the two steps, we obtain that .
Proof using the union of conjugacy classes definition of normality
Given: A group .
To prove: is a union of conjugacy classes in
.
Proof: The conjugacy classes form a partition of (arising from the equivalence relation of being conjugate), so
is their union.
Proof using the commutator definition of normality
Given: A group .
To prove: For every ,
.
Proof: This is direct from the fact that is closed under multiplication and inverses, and the commutator is defined in terms of these operations.