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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).

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    $\begingroup$ I'm actually puzzled: why isn't it trivially true (enumerate those finite groups $F_1,\dots,F_m$ of order $\le n$ admitting $A$ as quotient, and let $\theta(G)$ say: $G$ is isomorphic to one of the $F_i$)? $\endgroup$
    – YCor
    Commented Nov 18, 2020 at 21:22
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    $\begingroup$ If $H=\{h_1,\dots,h_k\}$ is a given finite group and $h_ih_j=h_{s(i,j)}$, a sentence $\theta(G)$ saying "$G$ is isomorphic to $H$" can be given as $\exists g_1,\dots,g_k: $ $(\forall g:(g=g_1)\vee (g=g_2)\dots\vee (g=g_k))$ $\wedge$ $(g_1\neq g_2)$ $\wedge$ $(g_1\neq g_3)$ $\dots$ $\wedge (g_{k-1}\neq g_k)$ $\wedge$ $(g_1g_2=g_{s(1,2)})$ $\wedge$ $(g_1g_3=g_{s(1,3)})$ $\dots$ $\wedge$ $(g_{k-1}g_k=g_{s(k-1,k)})$. $\endgroup$
    – YCor
    Commented Nov 19, 2020 at 8:06
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    $\begingroup$ Apart from what YCor wrote, how is $(L')$ not just a trivial reformulation of $(L)$? If $A$ is a quotient of $G$, then the order of $A$ divides the order of $G$, and by definition, $A$ is a quotient of $G$ iff there is a normal subgroup $M$ of $G$ such that $G/M\simeq A$. So why do you think it is weaker? $\endgroup$
    – Emil Jeřábek
    Commented Nov 19, 2020 at 9:20
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    $\begingroup$ Or is the concern just that $(L')$ requires $|G|=n$, while $(L)$ only requires $|G|\le n$? This is trivial to fix by just taking the disjunction of the sentences corresponding to each order up to $n$. $\endgroup$
    – Emil Jeřábek
    Commented Nov 19, 2020 at 9:25
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    $\begingroup$ @YCor I had a look at the paper, and the situation is actually quite bizarre: not only they do not need any particular property of their particular formula, they actually do not need the complicated statement of Lemma 1.1 at all! They only use it once in the proof of the main result, to show that two particular nonisomorphic finite groups can be distinguished by a sentence, but this can be shown more directly and much more easily using the observation in your comment above. $\endgroup$
    – Emil Jeřábek
    Commented Nov 19, 2020 at 10:51

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