Groups with no homomorphisms onto $\mathbb{Z}/p\mathbb{Z}$

Question

Does there exist any term for finite groups with no non-trivial homomorphisms into $\mathbb{Z}/p\mathbb{Z}$ for a fixed prime $p$, or any term related to this property (so that I could write "finite group with property X")?

Also, for a projective limit $G$ of some finite groups $G_i$ with no non-trivial homomorphisms into $\mathbb{Z}/p\mathbb{Z}$ can there exist a non-trivial homomorphism $G\to \mathbb{Z}/p\mathbb{Z}$? If it can, do any "non-trivial" assumptions on $G_i$ or $G$ guarantee the absence of homomorphisms of this sort (I don't want to assume that all subgroups of $G$ of finite index are closed or that $G$ is abelian)?

Upd. I am deeply grateful to YCor! So, my main question can be translated into: are projective limits of $p$-perfect finite groups $p$-perfect?

Answer 1

Yes, the answer ot this question is partially known. For example, there are products of finite $p$-perfect groups for $p = 2, 3$ which are not themselves $p$-perfect.

For each reduced word $w$ in $k$ letters one can define $w$-map $w_G: G^k \to G$, which just sends a $k$-tuple $(g_1, \dots, g_k)$ into the element $g_{w_1} g_{w_2} \dots g_{w_k}$. The group generated by the image of $w_G$ is called the verbal subgroup of $G$ for a word $w$, which we can denote as $V_G(w)$.

Define the width $m_G(w)$ of a word $w$ in group $G$ to be the minimal number such that any element of $V_G(w)$ is a product of $m_G(w)$ or fewer elements lying in the image of $w_G$, or $\infty$, if there's no such number.

Proposition. For a profinite (not necessarily finitely generated) group $P$, and a word $w$ the following conditions are equivalent:

1. $m_Q(w)$ is bounded over all continuous finite quotients $Q = P/N$;

2. $V_P(w)$ is closed in $P$;

3. $m_P(w)$ is finite.

For a short proof, you can look at proposition 4.1.2. in Dan Segal's "Words. Notes of Verbal Width in Groups". Original result is in B. Hartley, Subgroups of finite index in profinite groups, Math. Zeit. 168, (1979), 71–76.

---

So, if you somehow know that for your cofiltered diagram $\{G_i\}$ of finite groups $m_{G_i}(x^p)$ is bounded (which is the case if, for example, all $G_i$ are simple: Liebeck, O'Brien, Shalev, and Tiep, Products of squares in finite simple groups), then, indeed, the limit will be $p$-perfect. But how to obtain a counterexample?

One can cook up a sequence of finite groups which are 2-perfect, but $m_{G_i}(x^2)$ is not bounded (right now I'm not able to cook up an an explicit example of such, but something along the lines of iterated wreath products of alternating groups will work). Then one can look at $P := \prod G_i$.

I claim that $P$ is not 2-perfect. Why so? $V_P(x^2)$ is not closed, because width of $p$-power word is unbounded in finite continuous quotients; therefore, it cannot be equal to the whole group. On the other hand, $P/V_P(x^2)$ is nonzero, and is isomorphic to (usual, non-continuous) $H_1(P, \Bbb Z/2\Bbb Z)$.

You can replace 2 with 3 because we know how groups of exponent 3 look like – they are nilpotent, so have a $\Bbb Z/3\Bbb Z$ quotient. I'm not sure whether you can do similar things with 5; $G/V_P(x^5)$ could be a simple group, nobody really knows.