A way of getting proper normal subgroups of small index

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A way of getting proper normal subgroups of small index

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[note article incomplete]More detailed examples needed, as well as relevant definitions and general discussion of the idea.[/note]

[QUICK DESCRIPTION]

Given a group, it can be useful to know that there is a proper normal subgroup of small index, where 'small' is defined in terms of some finite invariant of the group.  To obtain such a result, it suffices to show that the group has a non-trivial action on a sufficiently small finite set.  These may arise from the internal structure of the group.

An action of a group $G$ on a finite set $X$ is nothing more or less than a homomorphism $\theta$ from $G$ to $\mathrm{Sym}(X)$, the group of permutations of $X$.  As $\mathrm{Sym}(X)$ has order $|X|!$, the image of $G$ under this homomorphism has order at most $|X|!$; in fact, it has order dividing $|X|!$, by Lagrange's theorem.  We thus obtain a normal subgroup of index dividing $|X|!$ as the kernel of this homomorphism.  The kernel is a proper normal subgroup if and only if the action of $G$ on $X$ is non-trivial.

[PREREQUISITES]

Basic group theory

[EXAMPLE]

Let $G$ be a group with a subgroup $H$ such that the index $|G:H|$ of $G$ in $H$ is finite.  Then there is a normal subgroup of $G$ contained in $H$ of index dividing $|G:H|!$ .

Proof: We have an action of $G$ by right multiplication on the set $X = \{Hg:g \in G \}$ of right cosets of $H$, that is, a homomorphism $\theta$ from $G$ to $\mathrm{Sym}(X)$.  Now let $N$ be the kernel of $\theta$; this is automatically a normal subgroup.  The cardinality of $X$ is precisely the index of $H$ in $G$, so $|\mathrm{Sym}(X)|=|G:H|!$ , and hence $|G:N|$ divides $|G:H|!$ .  The subset of $G$ that fixes the coset $H$ is precisely $H$ itself, so $N$ is contained in $H$.  Hence $N$ is the normal subgroup we require.

Indeed, it can be seen that the $N$ given above is in fact the largest normal subgroup of $G$ to be contained in $H$, known as the core of $H$ in $G$, in the sense that it contains all normal subgroups of $G$ that are contained in $H$.  This is because the stabiliser of the coset $Hg$ is the conjugate $g^{-1}Hg$ of $H$, and given any $K \leq H$ such that $K$ is normal in $G$, we have $K = K^g \leq H^g$.  So $K$ acts trivially on the right cosets of $H$, and is thus contained in $N$.

[EXAMPLE]

We say a subgroup $H$ of a group $G$ is characteristic if $H = H^\alpha$ for every automorphism $\alpha$ of $G$.  This is a stronger property than normality, and is useful primarily for the following reason: if $H$ is a subgroup of $G$, and $K$ is a characteristic subgroup of $H$, then every automorphism of $G$ that sends $H$ to itself induces an automorphism of $H$, and hence also sends $K$ to itself.  In particular, if $H$ is normal in $G$ then $K$ is normal in $G$, and if $H$ is characteristic in $G$ then $K$ is characteristic in $G$ (so the property of being a characteristic subgroup is transitive).

In addition, it can be said that 'any subgroup $K$ uniquely specified by a family $\mathcal{K}$ of characteristic subgroups is itself characteristic', even if $\mathcal{K}$ was chosen arbitrarily.  This is because any automorphism of $G$ will preserve $\mathcal{K}$ element-wise, so the specification of $K$ is unchanged by applying the automorphism.

Let $G$ be a group with normal (characteristic) subgroups $A$ and $B$, such that $A$ contains $B$ with $|A:B|$ finite, and such that the commutator $[G,A]$ of $G$ and $A$ is not contained in $B$.  Then $G$ has a proper normal subgroup of index dividing $|A:B|!$.

Proof: Since $A$ and $B$ are normal in $G$, we have an action of $G/B$ by conjugation on the quotient $A/B$.  Since $[G,A]$ is not contained in $B$ (which is the same as saying that $A/B$ is not in the centre of $G/B$), this action is non-trivial.  By mapping $G$ onto its quotient $G/B$ and composing this with the action of $G/B$ on $A/B$, we get a non-trivial homomorphism $\theta$ from $G$ to $\mathrm{Sym}(A/B)$.  The kernel $N$ of $\theta$ is therefore a proper normal (characteristic) subgroup of $G$ of index at most $|A:B|!$ .  If $A$ and $B$ are characteristic, then $N$ is characteristic, since it has been specified uniquely by our choice of $A$ and $B$.

If more is known about the structure of $A/B$, then the bound on the index of the  proper normal subgroup can be improved.  For instance, if it is known that all equivalence classes of elements of $A/B$ under the action of its automorphism group have size at most $k$, then $G/B$ must act on a single equivalence class non-trivially, giving a proper normal (characteristic) subgroup of index dividing $k!$.

For an example of how to use this in infinite group theory, suppose $A$ is a group with a characteristic subgroup $B$ of finite index that does not contain the derived subgroup $[A,A]$ of $A$ (this ensures that $A/B$ is non-abelian, so $A$ itself acts non-trivially on $A/B$ by conjugation).  Now suppose $G$ is any group containing $A$ as a normal (characteristic) subgroup, but of arbitrary index.  Then the argument above applies, and we obtain a proper normal (characteristic) subgroup $N$ of $G$ of finite index.  Moreover, if $A$ has a descending sequence of characteristic subgroups $A_i$ of finite index such that each factor $A_i/A_{i+1}$ is non-abelian, then we obtain a corresponding strictly descending sequence of normal (characteristic) subgroups of $G$.

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A way of getting proper normal subgroups of small index

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