Equivalence of conjugacy and commutator definitions of normality
This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
View a complete list of pages giving proofs of equivalence of definitions
Statement
Suppose is a subgroup of a group
and
is an element. Then the following are equivalent:
-
.
- The commutator
, where
is the subgroup generated by
.
If these equivalent conditions hold for all , then
is a normal subgroup of
. Note that the second condition on all
would translate to saying that
is contained in
.
Note that is a weaker condition than
for a particular
.
More on the proof techniques
The survey article manipulating equations in groups discusses similar proof techniques involving equations that deal with elements and subsets of groups.
Proof
(1) implies (2)
Given: Group , subgroup
,
such that
.
To prove: .
Proof: Since is a subgroup, it suffices to show that all the generators of
are in
, i.e., that
for every
. For this, note that:
By assumption, since
, and so the quotient
is also in
. This completes the proof.
(2) implies (1)
Given: Group , subgroup
,
such that
.
To prove: .
Proof: For any , we have:
By assumption, both and
are in
, hence
.