Here is the definition of a Sylow $p$-subgroup from Wikipedia:

For a prime number $p$, a Sylow $p$-subgroup (sometimes $p$-Sylow subgroup) of a group $G$ is a maximal $p$-subgroup of $G$, i.e., a subgroup of $G$ that is a $p$-group (so that the order of any group element is a power of $p$), and that is not a proper subgroup of any other $p$-subgroup of $G$.

The definition of a $p$-group is mentioned there in the brackets, but here is the definition as stated in its own article:

In mathematical group theory, given a prime number $p$, a $p$-group is a periodic group in which each element has a power of $p$ as its order: each element is of prime power order. That is, for each element $g$ of the group, there exists a nonnegative integer $n$ such that $g$ to the power $p^n$ is equal to the identity element.

My issue with these definitions concerns the trivial group. The group $\{\rm{id}\}$ has exactly one element of order $1$; since this is $p^0$ for any $p$, it follows that this group is a $p$-group for any prime $p$. In fact, this is the only group which is a $p$-group for more than one $p$, since if any element has an order greater than $1$, the prime that divides the order of this element must be the only $p$ for which $G$ is a $p$-group.

Given that the trivial group is a $p$-group, it also follows that nontrivial Sylow $p$-groups do not exist, because the trivial group is a subgroup of every group, so the only maximal $p$-group is the trivial group itself.

I am a bit bothered that no one seems to address this trivial case. Is the trivial group a $p$-group or not? Is it a Sylow $p$-group? Certainly the Sylow theorems are false under this definition, because there are no Sylow groups of size $p^n$ unless $n=0$.

  • 1
    $\begingroup$ What? How is that a problem with maximality? If there's a bigger group, then the trivial subgroup is not maximal. Additionally, a "power of $p$" is--in this context--understood to mean a positive one. $\endgroup$ – Adam Hughes Jul 20 '14 at 16:41
  • $\begingroup$ @AdamHughes I now see the error in my reasoning - I had the subset switched in my definition of a Sylow group, i.e. "if $H$ is a sylow group and $K\subset H$, then $K$ is not a $p$-group" (when it should read "if $H$ is a sylow group and $H\subset K\subseteq G$, then $K$ is not a $p$-group"). $\endgroup$ – Mario Carneiro Jul 20 '14 at 17:47
  • $\begingroup$ @AdamHughes Regarding "power of $p$" meaning positive power, I don't think that makes sense. If the definition of $p$-group was "all elements have order $p^n$ where $n>0$" then nothing would be a $p$-group because the identity element would always fail that, and if it was "all non-identity elements have order $p^n$ where $n>0$" then it's exactly the same as the usual definition. I'm still a little unsure of whether $\{\rm{id}\}$ is to be considered a $p$-group or not. What do the Sylow theorems say about the number of Sylow $p$-subgroups of $\{\rm{id}\}$? $\endgroup$ – Mario Carneiro Jul 20 '14 at 18:02
  • $\begingroup$ It doesn't really matter either way, everything is still true if you let in the identity subgroup. There's no contradictions in the theorems anywhere, it's just silly to talk about the trivial group. I've seen both definitions used, but even if you allow $\{e\}$ to be a subgroup, it doesn't cause any problems. The theorems say what they always say, they're just more vacuously true. $\endgroup$ – Adam Hughes Jul 20 '14 at 20:36

I may be misunderstanding your point, but is it about what a "proper" subgroup is? The usual definition is that a subgroup $H\leq G$ is a proper subgroup if $H\neq G$, and in particular, the trivial subgroup is a proper subgroup of $G$ unless $G$ itself is the trivial group.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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