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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, then is it true that $p$ divides $n+1$ ? I know that $p-1=\phi(p)|n$ but I cannot approach further . Please help . Thanks in advance

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    $\begingroup$ Just to point out that this makes sense. For example: $\mathbb Z_9 = \{ \bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5, \bar 6, \bar 7, \bar 8 \}$ is a p-group with $n=2$ elements of order $3$. $\endgroup$ – steven gregory Feb 25 '16 at 14:10
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    $\begingroup$ We have $n=t(p-1)$ where $t$ is the number of subgroups of order $p$. But $t \equiv 1 \pmod p$ by Sylow's Theorem, so $p$ divides $n+1$. $\endgroup$ – Derek Holt Feb 25 '16 at 14:36
  • $\begingroup$ @DerekHolt : I think Sylow's theorem say that no. of Sylow p-subgroups is of the form $pk+1$ ( which is $1$ here ) . Does it also say that no. of subgroups of order $p$ is of the form $ps+1$ ? $\endgroup$ – user228168 Feb 25 '16 at 14:46
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    $\begingroup$ Yes, Wielandt's proof gives that. See my answer to math.stackexchange.com/questions/479839/… According to that post, this result is due to Frobenius rather than to Sylow. $\endgroup$ – Derek Holt Feb 25 '16 at 15:07
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Every element of order $p$ is contained in the Sylow $p$ subgroup (this is a slight extension of Sylow II), so you want to know that the number of elements of order $p$ in a $p$-group, plus one, is divisible by $p.$ This is true for a cyclic group of order $p.$ Now, note that a $p$ group has non-trivial center, so has a central element of order $p,$ which should be enough for the induction step.

EDIT Here is a complete argument, along slightly different lines. As above, we only need to consider $p$-groups. Now, consider the set $X$ of elements of order $p$ in a $p$-group $G,$ and the conjugation action of $G$ on that set. We have a variant of the class equation: $$ |X| = n_Z + \sum |G|/C(x),$$ where the sum is over conjugacy classes of non-central elements, and $n_Z$ is the number of central elements of order $p.$ Every summand is divisible by $p,$ so we need to only consider $n_Z.$ Now, since the center is an abelian group, we need to show the result for abelian groups. By the fundamental theorem of abelian groups, we know that an abelian group is a product of cyclic groups. For cyclic group, the result is clear. Now, suppose $G = H_1 \times H_2,$ then, it is easy to see that $n(G) = n(H_1) + n(H_2) + n(H_1) n(H_2).$ Indeed, the elements of order $p$ in $G$ are those of the form $(1, x),$ where $x$ is of order $p$ in $H_2,$ and those of the form $(y, 1),$ with $y$ of order $p$ in $H_1,$ and those of the form $(x, y)$ (as above. Now use induction on the number of direct summands in $G.$

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  • $\begingroup$ If $f(p,G)$ denotes the no. of elements of $G$ of order $p$ , then is $f(p,G)=f(p,Z(G))+f(p,G/Z(G))$ ? $\endgroup$ – user228168 Feb 25 '16 at 14:22
  • $\begingroup$ @Saun Dev, no not true, take $G=Q$, the quaternion group of order $8$, and take $p=2$. $\endgroup$ – Nicky Hekster Feb 25 '16 at 21:06
  • $\begingroup$ @SaunDev See edit. $\endgroup$ – Igor Rivin Feb 26 '16 at 9:49
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This is the Exercise 24.59 in Gallian's Contemporary Abstract Algebra 8/e.

By Sylow 2nd Theorem, there exists only one Sylow $p$-subgroup $L$. For any subgroup $H$ of order $p$, by Sylow 1st Theorem, $H\leq L$. By Exercise 24.41, there exists a subgroup $N$ of order $p$, which is normal in $G$. By these two things, we can conclude that $N\lhd L$.

If $N\not\subseteq H\leq L$ and $|H|=p$, verify that $H/N=\{hN\mid h\in H\}$ is a subgroup of $L/N$. By Correspondence Theorem, there exists $K$ such that $N\leq K$ and $K/N=H/N$. Note that $|K/N|=|H/N|=p$ and $|K|=p^2$. By the Corollary of Theorem 24.2, $K$ 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 $L$ 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 $L$ correspond to a same subgroup $K$.

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

enter image description here

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

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