Order of a sixth root in group G This question originates from Chapter 10, H3 of the 2nd edition of A Book of Abstract Algebra by Charles C. Pinter.

Let $a$ denote an element of a group $G$.  Let $a$ have order 10.  If $a$ has a sixth root in $G$, say $a=b^6$, what is the order of $b$?

Here is what I think:

Given $\operatorname{ord}(a) = 10$ and $a=b^6$,
$\qquad a^{10} = e = (b^{6})^{10} = b^{60}$
Let $\operatorname{ord}(b) = x, x$ must divide $60 \implies x\in \{1,2,3,4,5,6,10,12,15,20,30,60\}.$

*

*$x=1: b = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$

*$x=2: b^2 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$

*$x=3: b^3 = e = b^6 = a$ but $\operatorname{ord}(a) \ne 1.$

*$x=4: b^4 = e = b^{12} = a^2$ but $\operatorname{ord}(a) \ne 2.$

*$x=5: b^5 = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$

*$x=6: b^6 = e = a$ but $\operatorname{ord}(a) \ne 1.$

*$x=10: b^{10} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$

*$x=12: b^{12} = e = a^2$ but $\operatorname{ord}(a) \ne 2.$

*$x=15: b^{15} = e = b^{30} = a^5$ but $\operatorname{ord}(a) \ne 5.$

*$x=20: b^{20} = e = b^{60} = a^{10}$.

*$x=30: b^{30} = e = a^5$ but $\operatorname{ord}(a) \ne 5.$

*$x=60: b^{60} = e = a^{10}.$
Hence $x \in \{20,60\}$.

So it seems $x=20$ as $20 < 60$.
Or is there a way to rule out $20$?
 A: $$\begin{align} o(b^6) & = \frac {o(b)} {\text {gcd}\ (o(b),6)}. \\ \implies o(b) & = 10 \cdot \text {gcd}\ (o(b),6). \\ \implies o(b) & = 10 \cdot \text {gcd}\ (10 \cdot \text {gcd}\ (o(b),6),6). \\ \implies o(b) & = 20 \cdot \text {gcd} (5 \cdot \text {gcd}\ (o(b),6) , 3). \\ \end{align}$$
Now if $3 \nmid o(b)$ then we have $o(b)=20.$ Otherwise if $3 \mid o(b)$ we have $o(b) = 60.$ So $o(b) = 20\ \text {or}\ 60.$
Observe that both the options are equally valid. Because for $n=20,60$ consider a generator $b$ of $\Bbb Z_n.$  Observe that $a=b^6$ has the same property as mentioned in the question. But for the first case we have $o(b)=20$ and for the second case we have $o(b)=60.$
A: Just realize in general

If $a$ has order $n$ and $a$ has a $k$th root $b$, then $b$ has order
  $\displaystyle\frac{nk}{l}$, where $n$ and $l$ are relatively prime.

Applying this to my original question where $n=10, k=6$.  So
$\qquad\displaystyle\operatorname{ord}(b) = \frac{10\cdot 6}{l}\qquad$ where $10$ and $l$ are relatively prime.
$\implies l \in \{1,3,7,9\}$


*

*$\displaystyle l=1: \operatorname{ord}(b) = \frac{60}{1} = 60$

*$\displaystyle l=3: \operatorname{ord}(b) = \frac{60}{3} = 20$

*$\displaystyle l=7: \operatorname{ord}(b) = \frac{60}{7}$ (not possible since it's not an integer)

*$\displaystyle l=9: \operatorname{ord}(b) = \frac{60}{9}$  (not possible since it's not an integer)


Hence $b$ has order of either $20$ or $60$.
$\Box$
