One of my research projects right now, in collaboration with a student,  involves a group-theoretic construction called a wreath product. Given two groups (in the algebraic sense), the wreath product of these groups is a new group constructed from them. In this post, I wish to discuss the order of elements in the wreath product and consider how this might obstruct groups from being isomorphic to a (non-trivial) wreath product.

Group actions and wreath products

We will have a group \(H\) that acts on a set \(\Omega\). That is, we have a function \(H\times\Omega \to \Omega\) where, if for each \(h\in H\) and \(\omega\in\Omega\) we write \(h\cdot\omega\) for the image of \((h, \omega)\) under the function, then the function satisfies:

\(1_H\cdot\omega = \omega,\quad\text{for all }\omega\in\Omega;\)

\((hh’)\cdot\omega = h\cdot(h’\cdot \omega),\quad\text{for all }h,h’\in H\ \text{and}\ \omega\in\Omega.\)

Equivalently, we may define \(\varphi: H \to \textrm{Sym}(\Omega)\) by setting \(\varphi(h)\), for each \(h\in H\), to be the bijection determined by \(\varphi(h)(\omega) = h\cdot\omega\) for \(\omega\in\Omega\). Being a group action means that \(\varphi\) is a well-defined group homomorphism.

Let \(G\) and \(H\) be groups, where \(H\) acts on a set \(\Omega\). For each \(h\in H\), this action associates to \(h\) an automorphism \(\alpha_h\) of the Cartesian product \(\prod_{\omega\in\Omega} G\) through the action on the indices. In other words, given \((g_{\scriptsize\omega})_{\omega\in\Omega}\), with each \(g_{\scriptsize\omega} \in G\), the automorphism associated to \(h\in H\) is given by

\(\alpha_h\left((g_{\scriptsize\omega})_{\omega\in\Omega}\right) = (g_{\scriptsize h\cdot\omega})_{\omega\in\Omega}.\)

With the automorphism on the Cartesian product \(\prod_{\omega\in\Omega} G\) associated to each element of \(H\) understood, the wreath product \(G \wr H\) of groups \(G\) and \(H\) is defined as the corresponding semi-direct product of \(\prod_{\omega\in\Omega} G\) and \(H\). Stated explicitly, elements of \(G \wr H\) have the form \((\bar g, h)\), where \(\bar g = (g_{\scriptsize\omega})_{\omega\in\Omega}\) is an element of the Cartesian product of \(G\) over \(\Omega\), and \(h\in H\); the operation on \(G \wr H\) is then

\( (\bar g, h)\cdot(\bar g’, h’)  =  ( \bar g \alpha_h(\bar g’), hh’ ). \)

Note that the inverse of \((\bar g, h)\) in the wreath product is equal to \(\left(\alpha_{h^{-1}}((\bar g)^{-1}), h^{-1}\right)\).

Letting \(\{1\}\) denote the trivial group, note that \(\{1\} \wr H \cong H\) for every group \(H\) (and a group action on any set \(\Omega\)). Let’s call this a trivial wreath product since you simply get the outer group that you began with. If \(H = \{1\}\), then, since the automorphism \(\alpha_1\) will be the identity, we get that \(G \wr H \cong \prod_{\Omega}G\) by the map \((\bar g, 1) \mapsto \bar g\).

Inhomogeneous wreath products

I want to consider a generalized version of the wreath product. Suppose that \(\Omega\) is partitioned into \(m\) subsets, \(\Omega = \Omega_1\cup\ldots\cup\Omega_m\), and there is a group action of \(H\) on \(\Omega_i\) for each \(i=1,2,\ldots,m\). Let \(G_1, G_2,\ldots, G_m\) be \(m\) groups. The corresponding inhomogeneous wreath product \((G_1,G_2,\ldots,G_m)\wr\wr H\) is defined to be the semi-direct product of \(P\) and \(H\) where

\(P = \prod_{i=1}^m\left(\prod_{\Omega_i}G_i\right).\)

Here, the automorphism of \(P\) associated to \(h\in H\) is that obtained by simultaneously using the \(m\) automorphisms on the product of each \(G_i\) over \(\Omega_i\): \(\alpha_h( (\bar g_1, \bar g_2, \ldots, \bar g_m) ) = (\alpha_h(\bar g_1), \alpha_h(\bar g_2), \ldots, \alpha_h(\bar g_m)). \)

If we have one group \(G\) such that \(G_i \cong G\) for all \(i=1,\ldots,m\), then we have that \((G_1,G_2,\ldots,G_m)\wr\wr H\) is isomorphic to \(G\wr H\), using the action of the outer group \(H\) on \(\Omega\).

For \(i=1,\ldots,m\), suppose that \(s_i = |\Omega_i|\) is the cardinality of \(\Omega_i\). Note that

\(|(G_1,G_2,\ldots,G_m)\wr\wr H| = |G_1|^{s_1}|G_2|^{s_2}\ldots|G_m|^{s_m}|H|.\)

Orders of Elements and Order of the Wreath Product

Given a finite group \(G\), consider the set \(\{n\in\mathbb N\ |\ \textrm{ord}(g)|n,\ \text{for all }g\in G\}\) and define \(n_G\) to be the minimum element of this set.¹

Lemma. Let \(G_1, G_2,\ldots, G_m\) be finite groups and, for convenience, write \(n\) for the least common multiple of the \(n_{G_i}\). Let \(H\) be a finite group and \(\Omega\) a set so that there is a partition \(\Omega = \Omega_1\cup\Omega_2\cup\ldots\cup\Omega_m\) such that \(H\) acts on \(\Omega_i\) for each \(i=1,2,\ldots,m\). If \(x \in (G_1,G_2,\ldots,G_m)\wr\wr H\) then \(x^{nn_H} = 1\).

Proof. Consider an element \(x = ((\bar g_1,\bar g_2,\ldots,\bar g_m), h)\) in \((G_1,G_2,\ldots,G_m)\wr\wr H\). Note that \(h^{n_H} = 1_H\).

Now, for each \(i=1,2,\ldots,m\) and \(\omega\in\Omega_i\), since all components of \(\bar g_i\) are in \(G_i\), there exist elements \(g_{i,\omega}^{(h)} \in G_i\) so that, writing \(\bar g_i^{(h)}\) for \((g_{i,\omega}^{(h)})_{\omega\in\Omega_i}\), we have \(x^{n_H} = ((\bar g_1^{(h)}, \bar g_2^{(h)}, \ldots, \bar g_m^{(h)}), 1_{H})\). Now, since \(1_{H}\) acts trivially on each \(\Omega_i\), we have that

\(x^{n_Hk} = (((\bar g_1^{(h)})^k, (\bar g_2^{(h)})^k, \ldots, (\bar g_m^{(h)})^k), 1_{H}).\)

If \(k\) is a multiple of \(n_{G_i}\) then \((g_{i,\omega}^{(h)})^k\) is the identity in \(G_i\), for any \(\omega\in\Omega_i\). Thus, in the Cartesian product, \((\bar g_{i}^{(h)})^n\) is the identity, for every \(i=1,2,\ldots,m\). This implies the result.        \(\blacksquare\)

The size of \(nn_H\) is generally quite small compared to \(|(G_1,G_2,\ldots,G_m)\wr\wr H|\). Indeed, if one of the \(G_i\) is the trivial group, then we can obtain the same group with a smaller value of \(m\) (or we simply have \(H\) if \(m=1\)). Hence, we may assume that \(|G_i| \ge 2\) for all \(i\).

Also, if \(|\Omega_i| = 1\) for some \(i\), then the construction produces a direct product with \(G_i\). This is another special case.

Outside of the above special cases, \(s_i \ge 2\) and \(|G_i|\ge 2\) for every \(i=1,2,\ldots,m\). Thus, since \(n_{G_i} \le |G_i|\) for each \(i\), we have that \(nn_H \le |G_1||G_2|\ldots|G_m||H|.\) Hence, for any \(x\in (G_1,G_2,\ldots,G_m)\wr\wr H\), we have that the ratio \(\frac{\textrm{ord}(x)}{|H|}\) is not more than the square root of \(|(G_1,G_2,\ldots,G_m)\wr\wr H|/|H|\).

Proposition. Suppose that \(A\) is a finite group and that there are groups \(G_1,G_2,\ldots,G_m,\) and \(H\) such that \(A \cong (G_1,G_2,\ldots,G_m)\wr\wr H\) is a non-trivial wreath product. Furthermore, let \(m\ge1\) be minimal so that this is the case. If there exists \(a\in A\) with \(\textrm{ord}(a) = \frac{|A|}p\) for a prime \(p\) then one of the following holds:

(i) \(A \cong G_1\times G_2\times \ldots\times G_m \times H\), or
(ii) either \(m=1\) and \(A \cong \mathbb Z_p\wr H\) or \(A\cong (\mathbb Z_p\wr H)\times G_2\times\ldots\times G_m\).

Proof. Using \(n\) for the least common multiple of \(n_{G_i}\), for \(i=1,2,\ldots,m\), the previous lemma implies that \(\textrm{ord}(a)\ \big|\ nn_H\). For each \(i\), since \(n_{G_i}\) is the least common multiple of orders of elements in \(G_i\) and \(|G_i|\) is also a common multiple, we have \(n_{G_i}\ \Big|\ |G_i|\). Putting this together, we have that

\(\textrm{ord}(a)\ \Big|\ |G_1||G_2|\ldots|G_m||H|.\)

Using \(s_i\) to denote \(|\Omega_i|\), the assumption that \(\textrm{ord}(a) = \frac{|A|}p\) means that

\(|G_1|^{s_1}|G_2|^{s_2}\ldots|G_m|^{s_m}|H|\ \Big|\ p|G_1||G_2|\ldots|G_m||H|\)

which implies \(|G_1|^{s_1-1}|G_2|^{s_2-1}\ldots|G_m|^{s_m-1}\ \Big|\ p\). If \(|G_i|=1\) for some \(i\) then, having \(H\) act on \(\Omega\setminus\Omega_i\), we have

\((G_1,\ldots,G_m)\wr\wr H \cong (G_1,\ldots, G_{i-1},G_{i+1},\ldots,G_m)\wr\wr H\).

This is not possible as our choice of \(m\) was minimal. Thus, we have \(|G_i|\ge 2\) for all \(i=1,\ldots,m\). Hence, at most one of \(s_1,s_2,\ldots,s_m\) is larger than 1 and if \(s_i > 1\) then \(s_i=2\) and \(|G_i| = p\).

If \(s_i = 1\) for all indices, then we are in case (i): \(A \cong G_1\times G_2\times \ldots \times H\). On the other hand, suppose that \(s_1 = 2\) (WLOG), and \(s_i = 1\) for \(i > 1\). We then have that \(G_1 \cong\mathbb Z_p\). Since \(|\Omega_i| = 1\) for \(i > 1\), the operation defined on the wreath product is identical to the direct product for the other groups \(G_i\). An isomorphism \(\psi\) that realizes \(A\cong (\mathbb Z_p\wr H)\times G_2\times\ldots\times G_m\) may be defined by

\(\psi\left((\bar g_1, g_2, \ldots, g_m), h\right) = \left((\bar g_1, h), g_2,\ldots,g_m\right).\)

 \(\blacksquare\)