# Characteristic submonoid of group not implies subgroup

## Statement

It is possible to have a group $G$ and a submonoid $M$ of $G$ such that $\sigma(M) = M$ for every automorphism $\sigma$ of $G$ such that $M$ is not a subgroup of $M$. (By submonoid, we mean a subset that is a monoid with the binary operation and identity element inherited from the whole group).

## Facts used

1. Characteristically simple and abelian implies characteristic in holomorph: If $A$ is a characteristically simple Abelian group, then $A$ is characteristic in its holomorph: the semidirect product of $A$ and $\operatorname{Aut}(A)$.
2. Automorphism group action lemma for quotients: Suppose $A$ is an abelian normal subgroup of a group $G$, and $\sigma$ is an automorphism of $G$ that restricts to an automorphism $\alpha$ of $A$ and descends to an automorphism $\sigma'$ of $G/A$. Then, if $\rho:G/A \to \operatorname{Aut}(A)$ denotes the map by the conjugation action of the quotient (note: this requires abelianness of the subgroup), then $\rho \circ \sigma' = c_\alpha \circ \rho$.

## Proof

### Example involving the affine group

Consider the group $G = GA(1,\mathbb{Q})$ (see general affine group:GA(1,Q)): the general affine group over the rational numbers. this group can be described concretely in many ways:

• It is the semidirect product of the additive group of rational numbers by the multiplicative group (i.e., it is the holomorph of the additive group of rational numbers).
• It is the group (under composition) of all linear maps $x \mapsto ax + b$ from $\mathbb{Q}$ to itself, with $a \in \mathbb{Q} \setminus 0$ and $b \in \mathbb{Q}$.
• It is the group of upper triangular $2 \times 2$ invertible matrices over the rationals, where both diagonal entries are equal.

Define $A$ as the following subgroup:

• It is the normal subgroup comprising the additive group of rational numbers: the base of the semidirect product.
• It is the subgroup comprising the translation maps: $x \mapsto x+ b$.
• It is the group of upper triangular $2 \times 2$ matrices with $1$s on the diagonal.

### A lemma=

Given: $G = GA(1,\mathbb{Q})$, $A$ is described as above, $\sigma$ is an automorphism of $G$.

To prove: $\sigma$ sends every coset of $A$ in $G$ to itself.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $A$ is a characteristic subgroup of $G$. Fact (1), and that the group of rationals is characteristically simple Description of $A$ in $G$. We can combine the given facts. Alternatively, we can explicitly compute that $A$ is the derived subgroup of $G$.
2 $\sigma$ restricts to an automorphism (say) $\alpha$ of $A$ and hence descends to an automorphism (say) $\sigma'$ of $G/A$. Step (1) direct from Step (1).
3 $\rho \circ \sigma' = c_\alpha \circ \rho$ where $\rho:G/A \to \operatorname{Aut}(A)$ is the map given by the action of $G$ on $A$ by conjugation (see quotient group acts on abelian normal subgroup), and where $c_\alpha$ denotes conjugation by $\alpha$. Fact (2) Step (2) Step-fact combination dorect.
4 $\rho:G/A \to \operatorname{Aut}(A)$ is an isomorphism and that $\operatorname{Aut}(A)$ is abelian The automorphism group of $\mathbb{Q}$ is $\mathbb{Q}^*$.
5 $\sigma'$ is the identity map on $G/A$. Steps (3), (4) By the abelianness of $\operatorname{Aut}(A)$ in Step (4), $c_\alpha$ is the identity map of $\operatorname{Aut}(A)$, so plugging into Step (3), $\rho \circ \sigma' = \rho$. Since $\rho$ is an isomorphism, $\sigma'$ is the identity map.
6 $\sigma$ sends every coset of $A$ in $G$ to itself. Steps (2), (5) This follows directly from Step (5) and by revisiting the definition of $\sigma'$ in terms of $\sigma$.

An immediate corollary of the above is that every union of cosets of $A$ in $G$ is a characteristic subset of $G$.

Based on the above, we can construct many examples of characteristic submonoids of $G$ that are not subgroups. The idea is to choose any submonoid of $\mathbb{Q}^* \cong G/A$, then take its inverse image in $G$ under the quotient map to get the characteristic submonoid of $G$. Some examples are below.

#### Contraction mapping submonoid

Define $M$ as the following submonoid (described in all the alternate descriptions):

• It is the subset of $G$ comprising those elements whose multiplicative group coordinate has modulus at most $1$.
• It is the set of all linear maps of the form $x \mapsto ax + b$ where $0 < |a| \le 1$ and $a,b \in \mathbb{Q}$.(in other words, it is those linear maps that are contractions).
• It is the set of all upper triangular $2 \times 2$ invertible matrices over the rationals where the two diagonal entries are equal and have absolute value at most $1$.

Clearly, $M$ is a submonoid of $G$ -- this follows from the fact that the set of $a$ for which $0 < |a| \le 1$ form a submonoid of $\mathbb{Q}^*$ under multiplication. On the other hand, $M$ is not a subgroup of $G$. By the preceding observations, $M$ is a characteristic submonoid of $G$ that is not a subgroup.

#### Nonnegative powers of 2 mapping submonoid

• It is the subset of $G$ comprising those elements whose multiplicative group coordinate is a nonnegative integer power of 2.
• It is the set of all linear maps of the form $x \mapsto 2^nx + b$ where $n$ is a nonnegative integer.
• It is the set of all upper triangular $2 \times 2$ invertible matrices over the rationals where the two diagonal entries are equal to each other and their value is of the form $2^n$, $n$ a nonnegative integer.

Clearly, $M$ is a submonoid of $G$ -- this follows from the fact that the set of nonnegative integer powers of 2 is a submonoid of $\mathbb{Q}^*$. On the other hand, $M$ is not a subgroup of $G$. By the preceding observations, $M$ is a characteristic submonoid of $G$ that is not a subgroup.

### Nilpotent group example

Further information: characteristic submonoid of nilpotent group not implies subgroup

It is possible to construct a nilpotent group (in fact, a nilpotent group in which every automorphism is inner) and a characteristic submonoid of the group (in fact, one that is contained in the center) that is not a subgroup.