Given a “free” finite group $G$, is the set of all its quotient groups closed under finite subdirect products?

Let $G$ be a finite group.

Let $n$ be an integer such that $G$ has the following property:

"$n$-Free property": $G$ can be generated by $n$ elements, and for any two generating $n$-tuples $(g_1,\ldots,g_n)$, $(g_1',\ldots,g_n')$ there is an automorphism $\sigma$ of $G$ such that $\sigma(g_i) = g_i'$.

Under this assumption, if $A,B$ are two quotients of $G$, then is every subgroup of $A\times B$ which projects onto $A$ and $B$ which is also generated by $n$ elements also a quotient of $G$?

Essentially, the idea behind my question is: Is the "Free property" sufficient to say that $G$ is a free pro-$C$ group of rank $n$ (in the sense of Ribes-Zalesski's characterization of free profinite groups in their book "Profinite Groups"), where here $C$ is the class of all finite subdirect products of quotients of $G$.

• @EricWofsey You're right of course. Apologies for the hasty formulation of the problem. – oxeimon Apr 10 '17 at 22:13
• I think you need $n$ to be the minimal number of generators, otherwise, no nontrivial finite group has the free property. – verret Apr 10 '17 at 22:22
• @verret Thanks, that is indeed the case I'm interested in. – oxeimon Apr 10 '17 at 22:24
• Do you really mean minimally generated by $n$ elements, or that $n$ is the minimum number of generators. For example, does the cyclic group of order $6$ have the free property? With your current definition, it does not, because it admits a minimal $2$-generating set, but not all $2$-generating sets are conjugate under the automorphism group. – verret Apr 10 '17 at 22:49
• @verret I think it should make more sense now. I would consider the cyclic group of order 6 to have the 1-free property, but not the 2-free property. – oxeimon Apr 10 '17 at 23:02

Okay, so I believe the answer is yes - here's a sketch:

Let $H$ be a subdirect product of $A$ and $B$, then by Goursat's lemma, there is a common quotient $C$ of $A,B$ such that $H$ can be written as the fiber product $H = A\times_C B$.

By assumption, $H$ is generated by $n$ elements, say $(h_1,\ldots,h_n)$, where $h_i = (a_i,b_i)$.

By assumption, there are surjections $f_A : G\rightarrow A$, and $f_B : G\rightarrow B$. Since $(a_1,\ldots,a_n)$ generate $A$, and $(b_1,\ldots,b_n)$ generate $B$, by Gaschutz's lemma, we may lift $(a_i),(b_i)$ (through $f_A,f_B$) to generating tuples $$(\tilde{a}_1,\ldots,\tilde{a}_n),\qquad\text{and}\qquad(\tilde{b}_1,\ldots,\tilde{b}_n)$$ of $G$. By the "free property", these two generating tuples are mapped to each other by some automorphism $\sigma\in\text{Aut}(G)$. Suppose $\sigma(\tilde{a}_i) = \tilde{b}_i$.

But then, consider the surjections $f_A : G\rightarrow A$ and $f_B\circ\sigma : G\rightarrow B$. Their product defines a map $$(f_A\times (f_B\circ\sigma)) : G\rightarrow H = A\times_C B$$ By construction, the image of $(\tilde{a}_1,\ldots,\tilde{a}_n)$ is precisely $(h_1,\ldots,h_n)$, which generate $H$, which proves that $H$ is a quotient of $G$.

As stated, no -- Take $A = B = G$ to be the trivial quotient of $G$ by $\{e\}$. Then $A \times B = G \times G$ is not a quotient of $G$ for most reasonable groups $G$.

• Ah, sorry, good point. I forgot a condition. I hope you don't mind me editing my question accordingly. – oxeimon Apr 10 '17 at 21:49
• The same answer is still a counterexample in many cases. For example, $A_5$ is $2$-generated, and so is $A_5\times A_5$, so just take $G=A=B=A_5$, and the subgroup $A_5\times A_5$ of itself. – verret Apr 10 '17 at 21:50
• @verret Thanks, you're right! Hopefully this will be my last edit. – oxeimon Apr 10 '17 at 21:58
• @verret But $A_5$ is not $2$-free. – Derek Holt Apr 11 '17 at 7:43
• @DerekHolt My answer and these comments were posted before the question was updated to include the $n$-free condition. – user263190 Apr 11 '17 at 16:20