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Let $n\in \Bbb N$ be fixed and $m\in \Bbb N$ be the least number such that there exists a group of order $m$ in which all groups of order $n$ can be (isomorphically) embedded.

Can we deduce $n!=m$?

Alexander Gruber
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  • I assume you mean "the smallest $m$ such that there exists a group of order $m$ into which all groups of order $n$ can be isomorphically embedded"? – Alexander Gruber Mar 08 '13 at 18:45
  • @AlexanderGruber: yes. the smallest group containing all.. I think sometimes $n!$ is the smallest number. –  Mar 08 '13 at 18:49
  • @julien: is $m=n!$ the smallest number such that there exists a group of order $m$ in which all groups of order n can be isomorphically embedded? –  Mar 08 '13 at 18:56
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    If $n$ is prime, doesn't $n=m$? So the answer to this question is "no"? – MartianInvader Mar 08 '13 at 22:10
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    For $n>2$ we always have $m < n!$. If $n$ is a prime power then you can use a Sylow $p$-subgroup of $S_{n!}$. Otherwise, all groups of order $n$ have a faithful permutation representation of degree at most $5n/6$. – Derek Holt Mar 09 '13 at 10:46
  • @DerekHolt: thanks. good result: any group of order $n$ acts faithfully on a set with not more than $\left \lfloor \frac{5}{6}n \right \rfloor$ elements. interesting. what's the action? can I find it in http://books.google.com/books?id=4QDpFN6k61EC&pg=PA1&source=gbs_toc_r&cad=3#v=onepage&q&f=true ? –  Mar 09 '13 at 15:15
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    I didn't say that was true for all $n$, I said it was true for all $n$ that are not powers of a prime. For example, the quaternion group $Q_8$ has no faithful action of degree less than 8. But 8 is a prime power. – Derek Holt Mar 10 '13 at 12:20

3 Answers3

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This is not known. In fact we don't even know how many groups of order $n$ exist for most $n$. A related open question asks what the smallest integer $s$ such that all groups of order $p^r$ embed in some group of order $p^s$.

However, I have reason to doubt that the number will be $m!$ for $m=p^e$. (This is not a proof, by the way, only a heuristic argument.) As in my link the growth rate of $p$-group isomorphism classes is $$p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3}.$$ Multiply this by $p^n$ to count the number of elements in each subgroup (which assumes generously they will be embedded without overlap, including the identity) and we get $$p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3+n}.$$ Comparing this to Stirling's approximation $$p^n!\sim \sqrt{2\pi p^n}\left(\frac{p^n}{e}\right)^{p^n}$$ you can see that the $p^n!$ will quickly become much too large to be the smallest group containing these isomorphism classes even if we were to insist that the groups were embedded completely separately.

In fact, we can use Stirling's approximation along with the upper bound in Gallagher (1967) to show that the number will almost never be $m!$ for large $m$. $$\lim_{m\to\infty}\frac{m^{cm^{2/3}\operatorname{log}(m)+1}}{\sqrt{2\pi m}\left(\frac{m}{e}\right)^m}=\lim_{m\to\infty}\frac{e^m m^{\frac{1}{2}-m+cm^{2/3} \operatorname{log}(m)}}{\sqrt{2 \pi }}= 0$$

So, I suspect that in general the number will be much lower than $m!$.

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Alexander Gruber
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  • Interesting, +1. I don't know much about finite groups. I'd be interested in an example of $n>2$ where $m=n!$ is actually the answer. – Julien Mar 08 '13 at 19:13
  • What if we changed it to require a group containing all groups of order at most $ n $? What if we changed it to "dividing $ n $"? – Tobias Kildetoft Mar 08 '13 at 19:19
  • For $p$-groups, you can go down to a Sylow $p$-subgroup of the symmetric group, which is smaller than the symmetric group itself. I would guess that you can't do much better than that, but I might have got that wrong! – Derek Holt Mar 08 '13 at 19:37
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    I just checked that there are groups of order 32, but none of order 16, that contain all 5 groups of order 8. But that is better than a Sylow 2-subgroup of $S_8$, which has order 128. – Derek Holt Mar 08 '13 at 19:47
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For a counterexample outside $n=p$ prime (where obviously $m=p$, since every order $p$ group is isomorphic to $\mathbb{Z}/p\mathbb{Z}$), consider $n=15$.

Then every order $15$ group is Abelian and is actually isomorphic to $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}=\mathbb{Z}/15\mathbb{Z}$.

So $m=15$ is the answer in this case.

Edit: of course, each time there is only one group of order $n$, the answer is $m=n$. See here for the beginning of a list of these cases.

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Julien
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No you can not deduce this. As a hint consider the prime numbers.

Baby Dragon
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  • in fact I wanted a more comprehensive answer for different n after thinking about this thread http://math.stackexchange.com/questions/323386/subgroups-of-symmetric-group/323408#323408. –  Mar 08 '13 at 19:23