Characteristic subgroup

In mathematics, particularly in the area of abstract algebra known as group theory, a characteristic subgroup is a subgroup that is mapped to itself by every automorphism of the parent group. Because every conjugation map is an inner automorphism, every characteristic subgroup is normal; though the converse is not guaranteed. Examples of characteristic subgroups include the commutator subgroup and the center of a group.

Definition
A subgroup $H$ of a group $G$ is called a characteristic subgroup if for every automorphism $φ$ of $G$, one has $φ(H) ≤ H$; then write $H char G$.

It would be equivalent to require the stronger condition $φ(H)$ = $H$ for every automorphism $φ$ of $G$, because $φ^{−1}(H) ≤ H$ implies the reverse inclusion $H ≤ φ(H)$.

Basic properties
Given $H char G$, every automorphism of $G$ induces an automorphism of the quotient group $G/H$, which yields a homomorphism $Aut(G) → Aut(G/H)$.

If $G$ has a unique subgroup $H$ of a given index, then $H$ is characteristic in $G$.

Normal subgroup
A subgroup of $H$ that is invariant under all inner automorphisms is called normal; also, an invariant subgroup.

Since $∀φ ∈ Inn(G)： φ[H] ≤ H$ and a characteristic subgroup is invariant under all automorphisms, every characteristic subgroup is normal. However, not every normal subgroup is characteristic. Here are several examples:
 * Let $Inn(G) ⊆ Aut(G)$ be a nontrivial group, and let $H$ be the direct product, $G$. Then the subgroups, $H × H$ and H  × {1$)$}}, are both normal, but neither is characteristic.  In particular, neither of these subgroups is invariant under the automorphism, ${1} × H$, that switches the two factors.
 * For a concrete example of this, let $(x, y) → (y, x)$ be the Klein four-group (which is isomorphic to the direct product, $$\mathbb{Z}_2 \times \mathbb{Z}_2$$). Since this group is abelian, every subgroup is normal; but every permutation of the 3 non-identity elements is an automorphism of $V$, so the 3 subgroups of order 2 are not characteristic.  Here $V$. Consider H = {e, a}}} and consider the automorphism, $V = {e, a, b, ab}$; then $T(e) = e, T(a) = b, T(b) = a, T(ab) = ab$ is not contained in $T(H)$.
 * In the quaternion group of order 8, each of the cyclic subgroups of order 4 is normal, but none of these are characteristic. However, the subgroup, {1, −1}}}, is characteristic, since it is the only subgroup of order 2.
 * If $H$ is even, the dihedral group of order $n$ has 3 subgroups of index 2, all of which are normal. One of these is the cyclic subgroup, which is characteristic.  The other two subgroups are dihedral; these are permuted by an outer automorphism of the parent group, and are therefore not characteristic.

Strictly characteristic subgroup
A strictly characteristic subgroup, or a distinguished subgroup, which is invariant under surjective endomorphisms. For finite groups, surjectivity of an endomorphism implies injectivity, so a surjective endomorphism is an automorphism; thus being strictly characteristic is equivalent to characteristic. This is not the case anymore for infinite groups.

Fully characteristic subgroup
For an even stronger constraint, a fully characteristic subgroup (also, fully invariant subgroup; cf. invariant subgroup), $2n$, of a group $H$, is a group remaining invariant under every endomorphism of $G$; that is,

Every group has itself (the improper subgroup) and the trivial subgroup as two of its fully characteristic subgroups. The commutator subgroup of a group is always a fully characteristic subgroup.

Every endomorphism of $G$ induces an endomorphism of $∀φ ∈ End(G)： φ[H] ≤ H$, which yields a map $G$.

Verbal subgroup
An even stronger constraint is verbal subgroup, which is the image of a fully invariant subgroup of a free group under a homomorphism. More generally, any verbal subgroup is always fully characteristic. For any reduced free group, and, in particular, for any free group, the converse also holds: every fully characteristic subgroup is verbal.

Transitivity
The property of being characteristic or fully characteristic is transitive; if $G/H$ is a (fully) characteristic subgroup of $End(G) → End(G/H)$, and $H$ is a (fully) characteristic subgroup of $K$, then $K$ is a (fully) characteristic subgroup of $G$.

Moreover, while normality is not transitive, it is true that every characteristic subgroup of a normal subgroup is normal.

Similarly, while being strictly characteristic (distinguished) is not transitive, it is true that every fully characteristic subgroup of a strictly characteristic subgroup is strictly characteristic.

However, unlike normality, if $H$ and $G$ is a subgroup of $H char K char G ⇒ H char G$ containing $H char K ⊲ G ⇒ H ⊲ G$, then in general $H char G$ is not necessarily characteristic in $K$.

Containments
Every subgroup that is fully characteristic is certainly strictly characteristic and characteristic; but a characteristic or even strictly characteristic subgroup need not be fully characteristic.

The center of a group is always a strictly characteristic subgroup, but it is not always fully characteristic. For example, the finite group of order 12, $G$, has a homomorphism taking $H$ to $H$, which takes the center, $$1 \times \mathbb{Z} / 2 \mathbb{Z}$$, into a subgroup of $K$, which meets the center only in the identity.

The relationship amongst these subgroup properties can be expressed as:
 * Subgroup ⇐ Normal subgroup ⇐ Characteristic subgroup ⇐ Strictly characteristic subgroup ⇐ Fully characteristic subgroup ⇐ Verbal subgroup

Finite example
Consider the group $H char G, H < K < G ⇏ H char K$ (the group of order 12 that is the direct product of the symmetric group of order 6 and a cyclic group of order 2). The center of $Sym(3) × $\mathbb{Z} / 2 \mathbb{Z}$$ is isomorphic to its second factor $$\mathbb{Z}_2$$. Note that the first factor, $(π, y)$, contains subgroups isomorphic to $$\mathbb{Z}_2$$, for instance $((1, 2)y, 0)$; let $$f: \mathbb{Z}_2<\rarr \text{S}_3$$ be the morphism mapping $$\mathbb{Z}_2$$ onto the indicated subgroup. Then the composition of the projection of $Sym(3) × 1$ onto its second factor $$\mathbb{Z}_2$$, followed by $G = S3 × $\mathbb{Z}_2$$, followed by the inclusion of $G$ into $S3$ as its first factor, provides an endomorphism of ${e, (12)}$ under which the image of the center, $$\mathbb{Z}_2$$, is not contained in the center, so here the center is not a fully characteristic subgroup of $G$.

Cyclic groups
Every subgroup of a cyclic group is characteristic.

Subgroup functors
The derived subgroup (or commutator subgroup) of a group is a verbal subgroup. The torsion subgroup of an abelian group is a fully invariant subgroup.

Topological groups
The identity component of a topological group is always a characteristic subgroup.