Skip to main content

a Lemma A lemma from Jarden's and Lubotzky's paper Elementary'Elementary equivalence of profinite groupsgroups'

fix link
Source Link
Emil Jeřábek
  • 47.1k
  • 4
  • 147
  • 208

I have a question about a reduction argument from Jarden's and Lubotzky's paper 'Elementary equivalence of profinite groups' in Lemma 1.1 on page 3:

Lemma 1.1: For each positive integer $n$ and each finite group $A$ of order at most $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ such that for every group $G$ of order at most $n$ the sentence $ \theta $ holds in $G$ if and only if $A$ is a quotient of $G$. $\tag{L}$

The proof begins with a reduction step I not understand:

Proof: It suffices to prove that for every positive integer $n$ and for every group $A$ of order $d$ dividing $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ with the following property: for every group $G$ of order $n$ the sentence $ \theta $ holds in $G$ if and only if $G$ has a normal subgroup $M$ such that $G/M \cong A$. [..] $\tag{L'}$

Assume we have proved (L'). Why the claim (L) of the lemma follows immediately from (L'); in other words why is sufficient to show only claim (L'), which is seemingly weaker as the claim (L)?

I have already asked identical question in MathStackExMathStackEx but a comment by Noah Schweber stresses another important viewpoint: That this lemma 1.1 is an elementary consequence of a much stronger statement about structures in context of finite languages. That's true, but that was not my original concern: my original concern is just about the logic of the proof itself: why (L') implies (L).

I have a question about a reduction argument from Jarden's and Lubotzky's paper 'Elementary equivalence of profinite groups' in Lemma 1.1 on page 3:

Lemma 1.1: For each positive integer $n$ and each finite group $A$ of order at most $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ such that for every group $G$ of order at most $n$ the sentence $ \theta $ holds in $G$ if and only if $A$ is a quotient of $G$. $\tag{L}$

The proof begins with a reduction step I not understand:

Proof: It suffices to prove that for every positive integer $n$ and for every group $A$ of order $d$ dividing $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ with the following property: for every group $G$ of order $n$ the sentence $ \theta $ holds in $G$ if and only if $G$ has a normal subgroup $M$ such that $G/M \cong A$. [..] $\tag{L'}$

Assume we have proved (L'). Why the claim (L) of the lemma follows immediately from (L'); in other words why is sufficient to show only claim (L'), which is seemingly weaker as the claim (L)?

I have already asked identical question in MathStackEx but a comment by Noah Schweber stresses another important viewpoint: That this lemma 1.1 is an elementary consequence of a much stronger statement about structures in context of finite languages. That's true, but that was not my original concern: my original concern is just about the logic of the proof itself: why (L') implies (L).

I have a question about a reduction argument from Jarden's and Lubotzky's paper 'Elementary equivalence of profinite groups' in Lemma 1.1 on page 3:

Lemma 1.1: For each positive integer $n$ and each finite group $A$ of order at most $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ such that for every group $G$ of order at most $n$ the sentence $ \theta $ holds in $G$ if and only if $A$ is a quotient of $G$. $\tag{L}$

The proof begins with a reduction step I not understand:

Proof: It suffices to prove that for every positive integer $n$ and for every group $A$ of order $d$ dividing $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ with the following property: for every group $G$ of order $n$ the sentence $ \theta $ holds in $G$ if and only if $G$ has a normal subgroup $M$ such that $G/M \cong A$. [..] $\tag{L'}$

Assume we have proved (L'). Why the claim (L) of the lemma follows immediately from (L'); in other words why is sufficient to show only claim (L'), which is seemingly weaker as the claim (L)?

I have already asked identical question in MathStackEx but a comment by Noah Schweber stresses another important viewpoint: That this lemma 1.1 is an elementary consequence of a much stronger statement about structures in context of finite languages. That's true, but that was not my original concern: my original concern is just about the logic of the proof itself: why (L') implies (L).

Source Link
user267839
  • 6k
  • 2
  • 11
  • 42

a Lemma from Jarden's and Lubotzky's paper Elementary equivalence of profinite groups

I have a question about a reduction argument from Jarden's and Lubotzky's paper 'Elementary equivalence of profinite groups' in Lemma 1.1 on page 3:

Lemma 1.1: For each positive integer $n$ and each finite group $A$ of order at most $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ such that for every group $G$ of order at most $n$ the sentence $ \theta $ holds in $G$ if and only if $A$ is a quotient of $G$. $\tag{L}$

The proof begins with a reduction step I not understand:

Proof: It suffices to prove that for every positive integer $n$ and for every group $A$ of order $d$ dividing $n$ there exists a sentence $ \theta $ of $\mathcal{L} \text{(group)} $ with the following property: for every group $G$ of order $n$ the sentence $ \theta $ holds in $G$ if and only if $G$ has a normal subgroup $M$ such that $G/M \cong A$. [..] $\tag{L'}$

Assume we have proved (L'). Why the claim (L) of the lemma follows immediately from (L'); in other words why is sufficient to show only claim (L'), which is seemingly weaker as the claim (L)?

I have already asked identical question in MathStackEx but a comment by Noah Schweber stresses another important viewpoint: That this lemma 1.1 is an elementary consequence of a much stronger statement about structures in context of finite languages. That's true, but that was not my original concern: my original concern is just about the logic of the proof itself: why (L') implies (L).