Why the group $\left$ has order a multiple of $p$? Let $G$ a finite group and $p$ a prime s.t. $p$ divide $|G|$. Let $g\in G$, $g\neq 1$ s.t. $p\nmid o(g)$ (where $o(g)$ denote the order of $g$). Consider the quotient $\bar G:=G/\left<g\right>$ and we suppose that $\bar G$ has an element of order $p$. Let denote $\bar h\in \bar G$ s.t. $o(\bar h)=p$. Let $h\in G$ s.t. $ \bar h=h\left<g\right>$. Now, it's written that $p\mid o(h)$ but I don't understand why. Any idea ?
 A: Let $m$ be the order of $h$ in $G$. Then $h^m=e$ in $G$. Therefore $\overline h^m=\overline e$ in $\overline G$. As $p$ is the order
of $\overline h$ in $\overline G$, then $p\mid m$.
A: We know that $o(\bar h) = p$. So, if $\bar h = h \langle g \rangle$, then $\bar h^{o(h)} = h^{o(h)} \langle g \rangle = e\langle g\rangle = \langle g\rangle$, so the order of $\bar h$, which is $p$, must divide $o(h)$.
A: There is a general result you can apply here:  

Let $G$ be a group and $N$ be a normal subgroup of $G$. For $a\in G$, $|Na|$ divides $|a|$.

To see this, note that $(Na)^{|a|}=N(a^{|a|})=Ne=N$ where $e$ is the identity. Hence $|Na|$ divides $|a|$ because $|Na|$ divides $k$ iff $(Na)^k=N$ (while this can be proven using division algorithm).
A: You need $\langle g\rangle$ normal in $G$ for the quotient to make sense...
If $\langle g\rangle \trianglelefteq G$, then $\bar h^n=e$ in the quotient, where $n=\mid h\mid$, by the first isomorphism theorem ($\frac {G}{\langle g\rangle}\cong \bar G$)  $\therefore p\mid n$.
