Supersolvable implies nilpotent derived subgroup: Difference between revisions

Latest revision as of 23:02, 5 July 2019

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., supersolvable group) must also satisfy the second group property (i.e., group with nilpotent derived subgroup)
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Statement

The derived subgroup of a supersolvable group is a nilpotent group. Moreover, the nilpotency class of the derived subgroup is bounded from above by the length of any normal series for the whole group where each of the quotient groups between successive members is a cyclic group.

Facts used

Derived subgroup centralizes cyclic normal subgroup
Derived subgroup is normal
Normality is strongly intersection-closed
Normality satisfies image condition
Second isomorphism theorem
Cyclicity is subgroup-closed
Derived subgroup satisfies image condition: Under a surjective homomorphism, the image of the derived subgroup equals the derived subgroup of the image.

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A supersolvable group $G$ with a normal series $1=N_{0}\leq N_{1}\leq \dots N_{c}=G$ with $N_{i}/N_{i-1}$ cyclic.

To prove: $[G,G]$ is nilpotent of class at most $c$ .

Proof: We will prove that the series:

$1=N_{0}\cap [G,G]\leq N_{1}\cap [G,G]\leq \dots \leq N_{c}\cap [G,G]=[G,G]$

is a central series for $[G,G]$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$[G,G]$ is normal in $G$	Fact (2)
2	$N_{i}\cap [G,G]$ is normal in $G$ for all $i$	Fact (3)	$N_{i}$ are all normal in $G$	Step (1)
3	$(N_{i}\cap [G,G])/(N_{i-1}\cap [G,G])$ is normal in $G/(N_{i-1}\cap [G,G])$ for all $1\leq i\leq c$	Fact (4)	$N_{i}$ are all normal in $G$	Step (2) (applied to both $i$ and $i-1$ )
4	$(N_{i}\cap [G,G])/(N_{i-1}\cap [G,G])$ is cyclic	Fact (6)	$N_{i}/N_{i-1}$ is cyclic	Step (2) (applied to $i-1$ )	By the second isomorphism theorem (fact (5)), this quotient is isomorphic to $N_{i-1}(N_{i}\cap [G,G])/N_{i-1}$ , which in turn is a subgroup of $N_{i}/N_{i-1}$ . Fact (6) therefore yields that it is cyclic.
5	$[G,G]/(N_{i-1}\cap [G,G])$ is the derived subgroup of $G/(N_{i-1}\cap [G,G])$	Fact (7)
6	$(N_{i}\cap [G,G])/(N_{i-1}\cap [G,G])$ is in the center of $[G,G]/(N_{i-1}\cap [G,G])$	Fact (1)		Steps (3), (4), (5)	$(N_{i}\cap [G,G])/(N_{i-1}\cap [G,G])$ is cyclic normal in $G/(N_{i-1}\cap [G,G])$ , and $[G,G]/(N_{i-1}\cap [G,G])$ is the derived subgroup. So fact (1) yields that $(N_{i}\cap [G,G])/(N_{i-1}\cap [G,G])$ is in the center of $[G,G]/(N_{i-1}\cap [G,G])$ .
7	The series is indeed a central series.			Step (6) (in light of step (2))

@@ Line 5: / Line 5: @@
 ==Statement==
-The [[fact about::derived subgroup]] of a [[fact about::supersolvable group]] is a [[fact about::nilpotent group]]. Moreover, the [[fact about::nilpotence class]] of the commutator subgroup is bounded from above by the length of any [[normal series]] for the whole group with cyclic quotients.
+The [[fact about::derived subgroup;3| ]][[derived subgroup]] of a [[fact about::supersolvable group;2| ]][[supersolvable group]] is a [[fact about::nilpotent group;3| ]][[nilpotent group]]. Moreover, the [[fact about::nilpotency class;3| ]][[nilpotency class]] of the derived subgroup is bounded from above by the length of any [[normal series]] for the whole group where each of the [[quotient group]]s between successive members is a [[cyclic group]].
 ==Facts used==
-# [[uses::Commutator subgroup centralizes cyclic normal subgroup]]
+# [[uses::Derived subgroup centralizes cyclic normal subgroup]]
-# [[uses::Commutator subgroup is normal]]
+# [[uses::Derived subgroup is normal]]
 # [[uses::Normality is strongly intersection-closed]]
 # [[uses::Normality satisfies image condition]]
 # [[uses::Second isomorphism theorem]]
 # [[uses::Cyclicity is subgroup-closed]]
-# [[uses::Commutator subgroup satisfies image condition]]: Under a surjective homomorphism, the image of the commutator subgorup equals the commutator subgroup of the image.
+# [[uses::Derived subgroup satisfies image condition]]: Under a surjective homomorphism, the image of the derived subgroup equals the derived subgroup of the image.
 ==Proof==
+{{tabular proof format}}
 '''Given''': A supersolvable group <math>G</math> with a normal series <math>1 = N_0 \le N_1 \le \dots N_c = G</math> with <math>N_i/N_{i-1}</math> cyclic.
@@ Line 29: / Line 31: @@
 is a [[central series]] for <math>[G,G]</math>.
-# <math>[G,G]</math> is normal in <math>G</math>: This is fact (2).
+{| class="sortable" border="1"
-# <math>N_i \cap [G,G]</math> is normal in <math>G</math> for all <math>i</math>: This follows from fact (3).
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
-# <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is normal in <math>G/(N_{i-1} \cap [G,G])</math> for all <math>1 \le i \le c</math>: This follows from fact (4).
+|-
-# <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is cyclic: By the second isomorphism theorem (fact (5)), this quotient is isomorphic to <math>N_{i-1}(N_i \cap [G,G])/N_{i-1}</math>, which in turn is a subgroup of <math>N_i/N_{i-1}</math>. Fact (6) therefore yields that it is cyclic.
+| 1 || <math>[G,G]</math> is normal in <math>G</math> || Fact (2) || || ||
-# <math>[G,G]/(N_{i-1} \cap [G,G])</math> is the commutator subgroup of <math>G/(N_{i-1} \cap [G,G])</math>: This is direct from fact (7).
+|-
-# <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is in the center of <math>[G,G]/(N_{i-1} \cap [G,G])</math>: By the previous three steps, <math>[G,G]/(N_{i-1} \cap [G,G])</math> is cyclic normal in <math>G/(N_{i-1} \cap [G,G])</math>, and <math>[G,G]/(N_{i-1} \cap [G,G])</math> is the commutator subgroup. So fact (1) yields that <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is in the center of <math>[G,G]/(N_{i-1} \cap [G,G])</math>.
+| 2 || <math>N_i \cap [G,G]</math> is normal in <math>G</math> for all <math>i</math> || Fact (3) || <math>N_i</math> are all normal in <math>G</math>|| Step (1) ||
+|-
-Thus, the series is indeed a central series.
+| 3 || <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is normal in <math>G/(N_{i-1} \cap [G,G])</math> for all <math>1 \le i \le c</math> ||  Fact (4) || <math>N_i</math> are all normal in <math>G</math> || Step (2) (applied to both <math>i</math> and <math>i - 1</math>) ||
+|-
+|  4 || <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is cyclic || Fact (6) || <math>N_i/N_{i-1}</math> is cyclic || Step (2) (applied to <math>i-1</math>) || By the second isomorphism theorem (fact (5)), this quotient is isomorphic to <math>N_{i-1}(N_i \cap [G,G])/N_{i-1}</math>, which in turn is a subgroup of <math>N_i/N_{i-1}</math>. Fact (6) therefore yields that it is cyclic.
+|-
+| 5 ||  <math>[G,G]/(N_{i-1} \cap [G,G])</math> is the derived subgroup of <math>G/(N_{i-1} \cap [G,G])</math> ||  Fact (7) || || ||
+|-
+| 6 || <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is in the center of <math>[G,G]/(N_{i-1} \cap [G,G])</math> || Fact (1) || || Steps (3), (4), (5) || <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is cyclic normal in <math>G/(N_{i-1} \cap [G,G])</math>, and <math>[G,G]/(N_{i-1} \cap [G,G])</math> is the derived subgroup. So fact (1) yields that <math>(N_i \cap [G,G])/(N_{i-1} \cap [G,G])</math> is in the center of <math>[G,G]/(N_{i-1} \cap [G,G])</math>.
+|-
+| 7 || The series is indeed a central series. || || || Step (6) (in light of step (2)) ||
+|}