I know this is a very common corollary of the class equation. And I know how to do it by using class equation.

But can you do it bu using group action, maybe find a nice set for G to act on?

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    $\begingroup$ But the class equation arises from an action of $G$ on a nice set. $\endgroup$ – Tobias Kildetoft Mar 19 '13 at 13:52
  • $\begingroup$ so is there any other way to do it without using class equation at all? $\endgroup$ – Akaichan Mar 19 '13 at 14:01
  • $\begingroup$ I am not aware of one that does not use the class equation of some disguised version of it. $\endgroup$ – Tobias Kildetoft Mar 19 '13 at 14:05

Let $G$ be a nontrivial finite $p$-group.

It is possible to prove without the class equation that in $G$, the number of subgroups of order $p$ is $\equiv 1 \mod{p}$. For example, we could use McKay's proof of Cauchy's theorem. From this it follows that the number of normal subgroups of order $p$ is also $\equiv 1 \mod{p}$, since the number of non-normal subgroups of order $p$ is $\equiv 0 \mod{p}$. This can be seen by having $G$ act on the non-normal subgroups by conjugation, every orbit has size $p^k$ for some $k > 0$.

Thus $G$ contains a normal subgroup $N$ of order $p$. By the normalizer-centralizer lemma, there exists a homomorphism $f: G \rightarrow \operatorname{Aut}(P)$ with $\operatorname{Ker}(f) = C_G(N)$. Now $\operatorname{Aut}(P) \cong \mathbb{Z}_p^*$ has order $p-1$ which is coprime to $p$, so $C_G(N) = G$ follows by $G / \operatorname{Ker}(f) \cong f(G)$ and Lagrange's theorem. Thus $N \leq Z(G)$ which proves that $Z(G)$ is not trivial.

Of course, proving this theorem this way is pretty silly. This is just a guess and my opinion, but I feel like the only easy way to prove this is with the class equation.


Consider the action of $G$ on itself by conjugation. Let $\chi_1,\ldots,\chi_n$ be the orbits of $G$ under this action. Since $G$ is a $p$-group, all orbits must have order $1$ or some power of $p$. But $$|G|=\sum\limits_{i=1}^n |\chi_i|$$ so since the LHS is divisible by $p$, the RHS is divisible by $p$ as well. Since all $|\chi_i|$ which aren't divisible by $p$ are $1$, the number of such $|\chi_i|$ must be divisible by $p$. So there are either no $\chi_i$ of cardinality $1$, or at least $p$. But $\chi_i$ of cardinality $1$ correspond precisely to elements of the center, and since $e$ is in the center there is at least $1$. So there are at least $p$.

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    $\begingroup$ Note that $|G|=\sum\limits_{i=1}^n |\chi_i|$ is the class equation. $\endgroup$ – spin Mar 19 '13 at 14:02
  • $\begingroup$ I consider it slightly misleading to not mention in the answer that this is really just using the class equation, since the question specifically asks for a solution without that. $\endgroup$ – Tobias Kildetoft Mar 19 '13 at 14:16
  • $\begingroup$ @TobiasKildetoft The OP didn't say so until a comment after I answered. It appeared to me that the OP was not aware of how the class equation arises. This really is the natural thing to do, so I'm not sure what else they're looking for. $\endgroup$ – Alex Becker Mar 19 '13 at 14:29

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