Characteristicity is quotient-transitive: Difference between revisions

Latest revision as of 04:40, 1 May 2022

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup)
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Statement

Property-theoretic statement

The subgroup property of being a characteristic subgroup satisfies the subgroup metaproperty of being quotient-transitive.

Statement with symbols

Suppose $H \leq K \leq G$ are subgroups such that $H$ is a characteristic subgroup of $G$ , and $K / H$ is a characteristic subgroup of $G / H$ . Then, $K$ is a characteristic subgroup of $G$ .

Related facts

Similar facts

Proof

Given: A group $G$ , subgroups $H \leq K \leq G$ such that $H$ is characteristic in $G$ , and $K / H$ is characteristic in $G / H$ . An automorphism $σ$ of $G$ .

To prove: $σ (K) \subseteq K$

Proof: First observe that $σ (H) = H$ , so $σ$ induces an automorphism on the quotient $G / H$ , by the rule $g H \mapsto σ (g) H$ . Call this automorphism $σ^{'}$ .

Then, $σ^{'}$ is an automorphism of $G / H$ . Since $K / H$ is characteristic in $G / H$ , $σ^{'} (K / H) = K / H$ . Thus, for any $g \in K$ , $σ^{'} (g H) \in K / H$ , and hence, unwrapping the definition, $σ (g) \in K$ . Thus, $σ (K) \subseteq K$ , completing the proof.

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 Suppose <math>H \le K \le G</math> are subgroups such that <math>H</math> is a [[characteristic subgroup]] of <math>G</math>, and <math>K/H</math> is a characteristic subgroup of <math>G/H</math>. Then, <math>K</math> is a characteristic subgroup of <math>G</math>.
+==Related facts==
+===Similar facts===
+* [[Normality is quotient-transitive]]
+* [[Full invariance is quotient-transitive]]
+* [[Strict characteristicity is quotient-transitive]]
 ==Proof==
-'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>
+'''Given''': A group <math>G</math>, subgroups <math>H \le K \le G</math> such that <math>H</math> is characteristic in <math>G</math>, and <math>K/H</math> is characteristic in <math>G/H</math>. An automorphism <math>\sigma</math> of <math>G</math>.
-'''To prove''': <math>K</math> is characteristic in <math>G</math>
+'''To prove''': <math>\sigma(K) \subseteq K</math>
-'''Proof''': We pick any automorphism <math>\sigma</math> of <math>G</math>, and want to show that <math>\sigma(K) = K</math>. For this, first observe that <math>\sigma(H) = H</math>, so <math>\sigma</math> induces an automorphism on the quotient <math>G/H</math>, by the rule <math>gH \mapsto \sigma(g)H</math>. Call this automorphism <math>\sigma'</math>.
+'''Proof''': First observe that <math>\sigma(H) = H</math>, so <math>\sigma</math> induces an automorphism on the quotient <math>G/H</math>, by the rule <math>gH \mapsto \sigma(g)H</math>. Call this automorphism <math>\sigma'</math>.
-Then, <math>\sigma'</math> is an automorphism of <math>G/H</math>. Since <math>K/H</math> is characteristic in <math>G/H</math>, <math>\sigma'(K/H) = K/H</math>. Thus, for any <math>g \in K</math>, <math>\sigma'(gH) \in K/H</math>, and hence, unwrapping the definition, <math>\sigma(g) \in K</math>. Thus, <math>\sigma(K) \subset K</math>. Since the same holds for <math>\sigma^{-1}</math>, we conclude that <math>\sigma(K) = K</math>, completing the proof.
+Then, <math>\sigma'</math> is an automorphism of <math>G/H</math>. Since <math>K/H</math> is characteristic in <math>G/H</math>, <math>\sigma'(K/H) = K/H</math>. Thus, for any <math>g \in K</math>, <math>\sigma'(gH) \in K/H</math>, and hence, unwrapping the definition, <math>\sigma(g) \in K</math>. Thus, <math>\sigma(K) \subseteq K</math>, completing the proof.