Question

If a prime $\mathit{p}$ divides the order of a finite group $\mathit{G}$, prove that the number of elements of order$\mathit{p}$ in$\mathit{G}$ is a multiple of $\mathit{p}\mathbf{-}\mathbf{1}$.

Short answer

It is proved that, number of elements of order$p$ in$G$ is a multiple of $p-1$.

Step-by-step solution

Determine that number of elements of order p in G is a multiple of p−1

Consider that, $G$is a finite group and $p$is the prime number. Here $p$divides the order of $G$.

If there is no element of order$p$in$G$ then the result is trivially as 0 as multiple of$p-1$.

Thus here at least one element in $G$of order $p$, say$x_1$.

Now considering a subgroup,

$H_1=⟨x_1⟩=\{e,x_1,{x_1}^2,\mathrm{...},{x_1}^{p-1}\}$

Here$p$ is the prime, every non-identical element in$H{}_1$ has order $p$. Hence a number of element in$H{}_1$ of order$p$ is$p-1$ .

Further simplification

Now taking element such as,$x_2,x_3,\mathrm{...},x_n\in G$ with property that order of each$x_j$ is$p$ and $x_j\in G/H_{j-1},j=2,\mathrm{...},n$, here,$H_j=⟨x_j⟩=\{e,x_j,{x_j}^2,\mathrm{...},{x_j}^{p-1}\}$

This is a cyclic group generated by $x_j$.

Now claim that for $i,j=1,2,\mathrm{...},n$with $i\ne j$,

$H_i\cap H_j=\left\{e\right\}$

Here$e$ is the identity element of$G$ .

Further simplification

Let’s consider that,$g\in H_i\cap H_j$ with$g\ne e$ for$i\ne j\in \{1,2,\mathrm{...},n\}$ then

$g={x_i}^k={x_j}^l$

For$0\le k,l\le p-1$ implies that,

$x_i={x_j}^{l+n-k+1}\in H_j$

This is not possible from the choice of $x_i,i=1,\mathrm{...},n$. Hence proved.