# If an element $c$ of order $d$ belongs to both $\langle a\rangle$ and $\langle b\rangle$, then $\langle a\rangle=\langle b\rangle=\langle c\rangle$?

If an element $c$ in a group $G$ belongs to both $\langle a\rangle$ and $\langle b\rangle$ (where $a$, $b$ belong to $G$), then if $|a|=|b|=|c|=d$, prove that $\langle a\rangle =\langle b\rangle =\langle c\rangle$.

$\langle \cdot\rangle$ denotes the cyclic group generated by $\cdot$ .

$Attempt$: Since $a$ and $b$ have the same order $d$ $\implies$ There exists a positive integer $p$ such that $a^p = c= b^p$.

And $c^{-1} = a^{d-p} = b^{d-p}$

For some reason, not able to figure out how do i prove that $\langle a\rangle =\langle b\rangle =\langle c\rangle$?

• Please use \langle and \rangle to get angle brackets $\langle$ and $\rangle$ in TeX. Both the symbols and the spacing are better that way. – Harald Hanche-Olsen Jan 2 '14 at 8:02

The problem with your attempt is that $p$ does not necessarily have to be the same for $a$ and $b$.
Hint: $\langle c \rangle \subseteq \langle a \rangle$ and $|\langle c \rangle|=|c|=|a|=|\langle a \rangle |$.
$$d=|c|=|a^p|=\frac{d}{\gcd(p,d)}\to\gcd(p,d)=1$$ $$a^1=a^{dq+pq'}=c^{q'}\to a\in\langle c\rangle$$