MHB Jordan-Holder Theorem for Groups .... Aluffi, Theorem 3.2

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0 ... I am currently focused on Chapter 4, Section 3: Composition Series and Solvability ...

I need help with an aspect of Aluffi's proof of the Jordan-Holder Theorem (Theorem 3.2, page 206) which reads as follows:

Theorem 3.2 and the early part of the proof read as follows:View attachment 4912
etc ... etc
In the above proof we read:

" ... ... We may assume $$G_1 \neq {G'}_1$$. Note that $$G_1 {G'}_1 = G$$: indeed, $$G_1 {G'}_1$$ is normal in $$G$$ and $$G_1 \subset G_1 {G'}_1$$ ... ... "
Question 1

Why does it follow from $$G_1 \neq {G'}_1$$ that $$G_1 {G'}_1 = G$$ ... ... ?Question 2

Further, how does it follow that $$G_1 {G'}_1$$ is normal in $$G$$ ... ?Question 3

Further yet, how does it follow that $$G_1 \subset G_1 {G'}_1$$ ... ... ?
I hope that someone can help ...

Peter

 

[Euge]

Hi Peter,

Question 1

Why does it follow from $$G_1 \neq {G'}_1$$ that $$G_1 {G'}_1 = G$$ ... ... ?

This question will be answered after I answer Questions 2 and 3. ;)

Question 2

Further, how does it follow that $$G_1 {G'}_1$$ is normal in $$G$$ ... ?

Recall that if $A$ and $B$ are normal subgroups of a group $G$, then $AB$ is normal in $G$. Apply this result with $A = G_1$ and $B = G_1^{'}$.

Question 3

Further yet, how does it follow that $$G_1 \subset G_1 {G'}_1$$ ... ... ?

Since $e\in G_1^{'}$, for all $g \in G_1$, $g = g\cdot e \in G_1 G_1^{'}$. Therefore, $G_1\subset G_1 G_1^{'}$.

Now, having answered the last two questions, I can answer Question 1. If $G_1 \neq G_1^{'}$, then there is some $g\in G_1^{'}$ such that $g\notin G_1$. So then $g = e\cdot g \in G_1G_1^{'}$, even though $g\notin G_1$. This means $G_1 \neq G_1G_1^{'}$. But since $G_1\subseteq G_1G_1^{'}$, we have $G_1 \subsetneq G_1 G_1^{'}$.

Consider the factor group $G/G_1$. It is a simple group by construction of the composition series, so it has no nontrivial normal subgroups. Using the lattice isomorphism theorem, we deduce that this is equivalent to there being no normal subgroups $N$ of $G$ such that $G_1 \subsetneq N \subsetneq G$. Since $N = G_1G_1^{'}$ is normal in $G$ and satisfies $G_1 \subsetneq N \subset G$, then by the previous statement, the only possibility is that $N = G$.

 

[Math Amateur]

Hi Peter,

This question will be answered after I answer Questions 2 and 3. ;)Recall that if $A$ and $B$ are normal subgroups of a group $G$, then $AB$ is normal in $G$. Apply this result with $A = G_1$ and $B = G_1^{'}$.

Since $e\in G_1^{'}$, for all $g \in G_1$, $g = g\cdot e \in G_1 G_1^{'}$. Therefore, $G_1\subset G_1 G_1^{'}$.

Now, having answered the last two questions, I can answer Question 1. If $G_1 \neq G_1^{'}$, then there is some $g\in G_1^{'}$ such that $g\notin G_1$. So then $g = e\cdot g \in G_1G_1^{'}$, even though $g\notin G_1$. This means $G_1 \neq G_1G_1^{'}$. But since $G_1\subseteq G_1G_1^{'}$, we have $G_1 \subsetneq G_1 G_1^{'}$.

Consider the factor group $G/G_1$. It is a simple group by construction of the composition series, so it has no nontrivial normal subgroups. Using the lattice isomorphism theorem, we deduce that this is equivalent to there being no normal subgroups $N$ of $G$ such that $G_1 \subsetneq N \subsetneq G$. Since $N = G_1G_1^{'}$ is normal in $G$ and satisfies $G_1 \subsetneq N \subset G$, then by the previous statement, the only possibility is that $N = G$.

Thanks so much for the help, Euge ... it is much appreciated ...

Just working through your post now ...

Thanks again,

Peter