Normal not implies characteristic
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about characteristic subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not characteristic subgroup|View examples of subgroups satisfying property normal subgroup and characteristic subgroup
Statement
It is possible to have a group and a normal subgroup
of
that is not a characteristic subgroup of
.
Related facts
Converse
Further information: Characteristic implies normal
The converse statement is indeed true. That is, every characteristic subgroup is normal.
Stronger formulations using the same examples
- Normal not implies characteristic in the collection of all groups satisfying a nontrivial finite direct product-closed group property
- Every nontrivial normal subgroup is potentially normal-and-not-characteristic
- Conjunction of normality with a nontrivially satisfied group property not implies characteristic
- Normality is not transitive: A normal subgroup of a normal subgroup need not be normal.
- Characteristic of normal implies normal: A characteristic subgroup of a normal subgroup is normal.
- Left transiter of normal is characteristic: If
is a subgroup of
such that whenever
is normal in
,
is normal in
, then
is characteristic in
.
- Direct factor not implies characteristic
Related group properties
There are some groups in which every normal subgroup is characteristic. Further information: Group in which every normal subgroup is characteristic
Proof
Example of a direct product
Let be any nontrivial group. Then consider
, viz., the external direct product of
with itself. The subgroups
and
are direct factors of
, and are hence both normal in
. Note also that they are distinct, since
is nontrivial.
However, the exchange automorphism:
exchanges the subgroups and
. Thus, neither
nor
is invariant under all the automorphisms, so neither is characteristic. Thus,
and
are both normal subgroups of
that are not characteristic.
Note that this example also shows that direct factor does not imply characteristic subgroup.
Particular cases of this example
- When
is the cyclic group of order two,
is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group. More generally, we can start with any nontrivial abelian group
.
Initial examples
Here are some examples where the ambient group is quite important and easy to understand:
Group part | Subgroup part | Quotient part | |
---|---|---|---|
Z2 in V4 | Klein four-group | Cyclic group:Z2 | Cyclic group:Z2 |
Here are some examples where the ambient group is somewhat more complicated:
Here are some examples where the ambient group is even more complicated:
Group part | Subgroup part | Quotient part | |
---|---|---|---|
D8 in D16 | Dihedral group:D16 | Dihedral group:D8 | Cyclic group:Z2 |
GAP implementation
Implementation of the generic example
Before using this generic example, you need to define for GAP, choosing any nontrivial group (double semicolons have been used here to suppress GAP's output for the first three commands, which depends on the specific choice of
-- you can use single semicolons instead).
gap> G := DirectProduct(H,H);; gap> H1 := Image(Embedding(G,1));; gap> H2 := Image(Embedding(G,2));; gap> IsSubgroup(G,H1); true gap> IsSubgroup(G,H2); true gap> IsNormal(G,H1); true gap> IsNormal(G,H2); true gap> IsCharacteristicSubgroup(G,H1); false gap> IsCharacteristicSubgroup(G,H2); false
References
Textbook references
- Topics in Algebra by I. N. HersteinMore info, Page 70 (Problem 7(a))
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347More info, Page 137 (Problem 6)