The number of elements of order $p$ in a $p$-group is $-1 \bmod p$

Question

I am reading a proof and it takes the following statement - but it is not immediate to me why it is true at all:

The number of elements of order $p$ ($p$ prime) in a $p$-group is $-1 \bmod p$.

Answer 1

if the group is abelian, then $a \to a^p$ is a homomorphism and the order of the kernel is a power of $p$ - as a subgroup of a $p$-group. however this kernel consists of the identity and all the elements of order $p$, hence the number of the latter is congruent to $-1 (\mod p)$.

since any $p$-group has a non-trivial center, the required result follows if it can be shown that the number of non-central elements of order $p$ is a multiple of $p$. but this follows by considering conjugacy classes, since every centralizer has order a positive power of $p$.

Answer 2

Here is another method.

Let $G$ be a $p$-group and $X=\{g\in G\mid |g|=p\}$ be the set of all the elements of order $p$. Let $G$ act on $X$ by conjugation. Note that \begin{eqnarray*} && |orbit(x)|=1\\ &\Leftrightarrow& orbit(x)=\{x\}\\ &\Leftrightarrow& \{g\cdot x=gxg^{-1}\mid g\in G\}=\{x\}\\ &\Leftrightarrow& \forall g\in G, ~gxg^{-1}=x\\ &\Leftrightarrow& x\in Z(G). \end{eqnarray*} Then by the class equation, \begin{eqnarray*} |X| &=&\sum_{\substack{x\in X\\|orbit(x)|=1}}|orbit(x)|+\sum_{\substack{i=1\\ |orbit(x_i)|\neq 1\\ orbit(x_i)\neq orbit(x_j)\text{ if }i\neq j}}^{n}|orbit(x_i)|\\ &=&\sum_{\substack{x\in X\\ x\in Z(G)}}1+\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)]\\ &=& |Z(G)\cap X|+\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)]. \end{eqnarray*} Then $$p\text{ divides }\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)]=|X|-|Z(G)\cap X|$$ (because $H$ is a $p$-group) and $$|X|\equiv |Z(G)\cap X|\pmod{p}.$$

(Tricky.) Verify that $(Z(G)\cap X)\cup \{e\}$ is a subgroup of $G$. Then by Lagrange's Theorem and $e\notin Z(G)\cap X$, $|Z(G)\cap X|+1=|(Z(G)\cap X)\cup \{e\}|$ divides $|G|$. It follows that $$|Z(G)\cap X|\equiv -1\pmod{p}.$$

Answer 3

This question is from Rotman's An Introduction to the Theory of Groups 4/e, p.75, Lemma 4.7 or Isaacs's Finite Group Theory, p.7, 1A.8.(b). There is another exercise in Gallian's Contemporary Abstract Algebra 8/e. Which is essentially the same as this problem. See exercise 24.59. Here is the author's proof.

Corollary of Theorem 4.4. Let $G$ be a group and $|G|=n$. If $d\mid n$, then the number of elements of order $d$ is $\phi(d)k$ for some $k\in \Bbb{Z}$, where $\phi$ is the Euler's totient function.

We prove your assertion. Let $G$ be a $p$-group and $|G|=p^n$. By Lagrange's Theorem, the order of every element in $G$ is a power of $p$. Thus, \begin{eqnarray*} \#\{g\in G:|g|=p\} &=&|G|-\#\{g\in G:|g|=p^n\}-\#\{g\in G:|g|=p^{n-1}\}\\ &-&\cdots -\#\{g\in G:|g|=p^2\}-\#\{g\in G:|g|=1\}\\ &=&p^n-\phi(p^n)k_n-\phi(p^{n-1})k_{n-1}-\cdots -\phi(p^2)k_2-1\\ &=&ps-1. \end{eqnarray*}

Exercise 24.59. Suppose $G$ is a finite group and $p$ is a prime that divides $|G|$. Let $n$ denote the number of elements of $G$ that have order $p$. If the Sylow $p$-subgroup of $G$ is normal, prove that $p$ divides $n+1$.

Answer 4

Here is my proof. The idea was obtained by plenty observations.

Let $G$ be a $p$-group and $|G|=p^n$. By the exercise II.5.3 in Hungerford's Algebra, there exists a normal subgroup $N$ of order $p$ in $G$.

If $N\neq H\leq G$ and $|H|=p$, that is, $H$ is another subgroup of order $p$, verify that the set $S=\{hN\mid h\in H\}$ is a subgroup of $G/N$. By Correspondence Theorem, there exists $K$ such that $N\leq K$ and $K/N=S$. Note that $|K/N|=|S|=p$ and $|K|=p^2$. By the exercise II.5.13 in Hungerford's Algebra, $K$ is abelian. Write $K/N=\langle kN\rangle$, where $k\notin N$.

Since $$\langle kn_i\rangle/N \ni(kn_i)^mN \stackrel{K\text{ is abelian}}{=}k^m n_i^mN =k^m N =(kN)^m \in \langle kN\rangle =K/N, $$ we have $\langle kn_i\rangle/N=K/N$ for each $n_i\in N=\{n_1, n_2, ...,n_p\}$. Note that $|\langle kn_i\rangle/N|=|K/N|=p$. So $\langle kn_i\rangle$ is a subgroup of order $p$. Hence, $\langle kn_1\rangle, \langle kn_2\rangle, ..., \langle kn_p\rangle$ are all distinct subgroups of $G$ whose order is $p$ such that $\langle kn_i\rangle/N=K/N$. The $p$ subgroups $\langle kn_1\rangle, \langle kn_2\rangle, ..., \langle kn_p\rangle$ of $G$ correspond to a same subgroup $K$.

Similarly, except $N$, every $p$ subgroups of order $p$ in $G$ correspond to a same subgroup of order $p^2$. Let $s$ be the number of subgroups of order $p$ in $G$. Then $p\mid (s-1)$. Let $r$ be the number of elements of order $p$ in $G$. Note that $s=\frac{r}{p-1}$, as the following figure indicates.

Therefore, \begin{eqnarray*} p\mid (s-1)=\left(\frac{r}{p-1}-1\right) &\Rightarrow& p(p-1)\mid r-(p-1)\\ &\Rightarrow& p\mid r-p+1\\ &\Rightarrow& p\mid r+1 \end{eqnarray*}

Answer 5

You can get it as a corollary of the more general following:

Theorem. Let $p$ be a prime, $G$ a group of order $|G|=p^\alpha m$, where $\alpha\ge 1$ and $p\nmid m$. Denoted with $n_k:=\#$ of subgroups of order $p^k$, we get: $$n_k\equiv 1\pmod p$$ for every $k=1,\dots,\alpha$.

Proof: see this post.

Corollary. For $k=1$, the Theorem states that the number of subgroups of order $p$ is $n_1=lp+1$ for some nonnegative integer $l$. Each of them comprises $p-1$ elements of order $p$, which therefore are overall $N=(p-1)(lp+1)\equiv$ $-1\pmod p$.