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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 $?

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

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$$d=|c|=|a^p|=\frac{d}{\gcd(p,d)}\to\gcd(p,d)=1$$ $$a^1=a^{dq+pq'}=c^{q'}\to a\in\langle c\rangle$$

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