MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm wondering about the following 2 generalizations of Cayley's Theorem (every group embeds in a symmetric group). If these are known to be true/false, references would be appreciated.

1) (Weak Version) Given any finite collection of (not necessarily distinct) finite groups, can we embed them simultaneously in a (finite) group so that they have pairwise-disjoint intersection (i.e., intersect only at the identity)?

2) (Strong Version) Given an arbitrary set of arbitrary groups, can we embed them simultaneously in some group so that they have pairwise-disjoint intersection? It seems like this version may run into set-theoretic difficulties; if so an explanation/reference for those would also be welcome.

Note of course that if such a group exists for a given collection, then we can embed it in a symmetric group by Cayley's Theorem (so these are indeed generalizations).

Edit: To make it more interesting / rule out the obvious answer pointed out in the comments, can we arrange the embeddings so that the groups' normalizers are pairwise disjoint?

share|cite|improve this question
Regarding your question 1, you can consider the direct product of the finite groups. – François Brunault Sep 26 '12 at 10:05
I don't understand the difficulty. The product of the set of groups exists (he didn't say class). – Todd Trimble Sep 26 '12 at 11:42
To amplify Todd's comment: If you wanted to do this for a proper class of (non-trivial) groups, then the "group" you embed them into will obviously need to be a proper class. But as long as you're dealing with a set of groups, no set-theoretic issues arise. – Andreas Blass Sep 26 '12 at 11:55
You can do much better than taking the direct product: you can take the direct product of $S_{G_i}$ (permutation group on the underlying set of the $G_i$) and view it as a subgroup of $S_X$, where $X$ is the disjoint union of the $G_i$; even better, you can identify the identities of the groups, so that $X= \{e\} \coprod(G_i\setminus\{1\})$, where $\coprod$ is the disjoint union. This embeds the family into a symmetric group on $1+\sum(|G_i|-1)$ letters. – Arturo Magidin Sep 26 '12 at 16:30
@Jon : Re your new question, the free product of the $G_i$'s will work. The normalizer of $G_1$ inside $G_1 * G_2$ is equal to $G_1$ (except in the trivial case $G_1=\{1\}$). This can be proved using reduced word decomposition in $G_1 * G_2$. – François Brunault Sep 27 '12 at 8:28
up vote 3 down vote accepted

So the question now is, given two nontrivial finite groups $G,H$, can we embed them both in a finite group $X$ such that the normalizers of $G$ and $H$ in $X$ intersect trivially?

I think we can do that as follows. Suppose that we can find a module $V$ for $G \times H$ over a finite field, such that neither $G$ nor $H$ fix any nonzero vectors $v \in V$, and such that there exists a vector $v \in V$ such that no nontrivial element of $G \times H$ stabilizes $v$. Then we let $X$ be the semidirect product of $V$ by $G \times H$. Because of the first property, the normalizers of $G$ and $H$ in $X$ have trivial intersection with $V$, so they are both equal to $G \times H$. The second property ensures that the complements $G \times H$ and $(G \times H)^v$ intersect trivially. So the subgroups $G$ and $H^v$ have the desired property, because their normalizers in $X$ are $G \times H$ and $(G \times H)^v$.

We can contruct such a module $V$ as follows. Choose a prime $p$ not dividing $|G|$ or $|H|$ and let $V_1$ and $V_2$ be the deleted permutation modules for the regular permutation representations of $G$ and $H$ over ${\mathbb F}_p$. (So $V_1$ and $V_2$ have dimensions $|G|-1$ and $|H|-1$.) These have vectors $v_1$ and $v_2$ that are not stabilised by any nontrivial element of $G$ and $H$, respectively. Now let $V$ be the $G \times H$ module $V_1 \otimes V_2$ and let $v =v_1 \otimes v_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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