4
$\begingroup$

Is every finite $p$-group an epimorphic image of a fibered product of two finite $p$-groups which can be generated by $2$ elements?

$\endgroup$
2
  • $\begingroup$ Wouldn't that limit the number of generators for such a group? $\endgroup$
    – Steve D
    Mar 22, 2015 at 16:13
  • 2
    $\begingroup$ You can obtain all finite abelian $p$-groups $A$: indeed embed $A$ in the center $Z$ of a 2-generated $p$-group $G$ and consider the fibered product of $G$ with itself over $Z$. Then its quotient by the diagonal of $G$ in $G\times G$ is isomorphic to the quotient of $Z\times Z$ by its diagonal and hence admits $A$ as a quotient. $\endgroup$
    – YCor
    Mar 22, 2015 at 16:44

1 Answer 1

6
$\begingroup$

No, you cannot obtain all finite $p$-groups in this way; you even miss some $p$-groups of nilpotency length 2 (although you have all abelian $p$-groups as I mentioned in a comment).

Lemma: Let $G$ be a fibered product of $G_1$ and $G_2$. Let $L$ be a subgroup of $G$ both of whose projection on $G_1$ and $G_2$ are onto, and let $N$ be the normal subgroup of $G$ generated by $L$. Then $G/N$ is abelian.

Proof: Let $G$ be fibered product of $G_1\stackrel{f_1}\to H\stackrel{f_2}\leftarrow G_2$. Let $K_i$ be the kernel of $f_i$. Let $p$ be the projection $G\to G/N$. It follows from the definition of $N$ that $p(K_1\times K_2)=G/N$. So to show that $G/N$ is abelian, it is enough to show that $p(K_1)$ is abelian, i.e., $p([K_1,K_1])$ is abelian. If $u,v\in K_1$, let $w$ be an element in $G_2$ such that $(v,w)\in L\subset N$. Then $$p([u,v],1)=p([(u,1),(v,w)])=[p(u,1),p(v,w)]=[p(u,1),1]=1.\qquad\Box$$

To apply the lemma in your context, assume that $G_1$ and $G_2$ are generated by $k_1$ and $k_2$ elements (you have $k_1=k_2=2$ but the argument is more general). Then we can find $L$ as in the lemma, generated by $k=k_1+k_2$ elements (actually we could even do $k_1+k_2-s$ where $s$ is the minimal number of generators of $H$). Hence there exist $s$ elements in $G$ ($s=4$ in your case) such that the quotient of $G$ by the normal subgroup generated by these $s$ elements is abelian.

It is thus enough to find a $p$-group in which this does not hold. For instance, if we take a free group on $c$ generators in the variety of 2-nilpotent groups of exponent $p$ (exponent 4 if $p=2$) then it works: such a group is thus not (a quotient of) a fibered product of 2 2-generated groups if $c$ is large enough (namely $c\ge 6$), because its quotient by the normal subgroup generated by 4 elements is never abelian.

Edit: here's a proof of the latter fact (for odd prime $p$): the truncated Baker-Campbell-Hausdorff formula yields an equivalence between the categories of 2-nilpotents Lie algebras over $\mathbf{Z}/p\mathbf{Z}$ and 2-nilpotents groups of exponent dividing $p$. In the free 2-nilpotent Lie algebra $\mathfrak{f}$ on generators $X_1\dots,X_n$ (and over any field), the ideal generated by $X_1,\dots,X_k$ is spanned by $X_1,\dots,X_k$ and $[X_i,X_j]$ for $i\le k$ and $j\le n$; its intersection $W_k$ with $[\mathfrak{f},\mathfrak{f}]$ thus has dimension $f(k)=k(k-1)/2+k(n-k)$. Note that by construction, $f(k-1)+1\le f(k)$ for all $1\le k\le n-1$, because $[X_{k},X_{k+1}]\in W_k\smallsetminus W_{k-1}$. In particular, $f(k)<f(n)=\dim(\mathfrak{f})$ whenever $k\le n-2$, and $f(k)+s\le f(k+s)$ whenever $0\le k\le k+s\le n$. Now consider a subspace $V$ of dimension $m$ in $\mathfrak{f}$. Let $s$ be the dimension of its intersection with the derived subalgebra $[\mathfrak{f},\mathfrak{f}]$ and $k=m-s$, so $k$ is the dimension of the projection to the abelianization. Then up to an automorphism of the $\mathfrak{f}$, we can suppose that $V$ contains $X_1,\dots,X_k$, and hence if generated by those elements and $s$ elements of the derived subalgebra. So the ideal generated by $V$ has dimension $\le f(k)+s$, which is $\le f(k+s)=f(m)$ as we have seen. Thus if the ideal generated by $V$ contains $[\mathfrak{f},\mathfrak{f}]$ then $f(m)=n$, which implies $m\ge n-1$; in other words, the quotient of $\mathfrak{f}$ by any ideal generated by $\le n-2$ elements is non-abelian (of course it's optimal since killing $n-1$ canonical generators yields a 1-dimensional Lie algebra).

All these facts carry over to the group, because the equivalence of categories is the identity on the underlying sets: namely given odd $p$ and an 2-nilpotent $p$-group, there is a canonically defined addition and Lie algebra law yielding the group law by the Baker-Campbell-Hausdorff formula $(gh=g+h+(1/2)[g,h])$ and Lie subalgebras (resp ideals) coincide with subgroups (resp. normal subgroups), so the minimal number of generators in either sense is the same as a group or subalgebra, resp. as a normal subgroup or ideal. Of course another approach is to convert the previous proof to the group setting, which should not be hard, and would allow to encompass the case of 2-groups.

$\endgroup$
4
  • $\begingroup$ How do you know that in the free group on $6$ generators in the variety of $2$-nilpotent groups of odd exponent $p$ the quotient by the normal subgroup generated by $4$ elements is not abelian? $\endgroup$
    – Pablo
    May 20, 2015 at 8:02
  • 1
    $\begingroup$ @Pablo I edited to include a proof. $\endgroup$
    – YCor
    May 20, 2015 at 14:35
  • $\begingroup$ This is a very interesting argument! However I suspect that one can (more easily) find a finite $p$-group with $6$ generators in which the quotient by the normal subgroup generated by any $4$ elements is not abelian. Actually, this reduces to showing that a pro-$p$ group with $6$ generators and $4$ (profinite) relations is not abelian (this is the same as saying that the free pro-$p$ group on $6$ elements mod a normal subgroup generated by any $4$ elements is not abelian). Still I am not sure how to prove this seemingly simple claim. Do you happen to have an idea? $\endgroup$
    – Pablo
    May 20, 2015 at 17:29
  • $\begingroup$ It expect that any proof boils bown to the case of 2-nilpotent $p$-groups in which the derived subgroup has exponent $p$. So you won't find any simplification by any other mean; you can just modify the argument to avoid the Lie algebra stuff which is most likely unnecessary. $\endgroup$
    – YCor
    May 20, 2015 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.