Index of $\langle a^4\rangle$ in the group $\langle a\rangle$ Let $a$ be an element of order 30 in a group $G$. What is the index of $\langle a^4\rangle$ in the group $\langle a \rangle$? 
The answer is 2, however I have no idea on how to obtain that answer. Maybe try to apply Lagrange's theorem?
 A: See what the group $ \langle a^4 \rangle$ is: $e, a^4, a^8, a^{12}, a^{16}, a^{20}, a^{24}, a^{28}, a^{32} = a^{2}, a^6, a^{10}, a^{14}, a^{18}, a^{22}, a^{26}, a^{30}=e$
You get all the elements with even power. That is you get half of the elements. Hence the index is two (you have two cosets: $ \langle a^4 \rangle$ and $a \langle a^4 \rangle$ ).
A: Hint: Let $o(a)$ denotes the order of $a$, then for any positive integer $n$ we have
$$o(a^n)=\frac{o(a)}{\gcd(o(a),n)}.$$
A: The simplest way I know is to list the elements of
$\langle a^4  \rangle$.  We have:  $a^4, a^8, a^{12}, a^{16}, a^{20}, a^{24}, a^{28}$, all distinct since their exponents  are less than $30$.  From there, we have $a^{32} = a^2$, since $a^{30} = e$, the group identity, and from there $a^6, a^{10}, a^{14}, a^{18}, a^{22}, a^{26}, a^{30} = e$; a total of $15$ elements in all.  So $\mid \langle a^4 \rangle \mid = 15$.  Since $\mid \langle a \rangle \mid  = 30$, by Lagrange's theorem we have $[\langle a \rangle : \langle a^4 \rangle ] = 30/15 = 2$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: The subgroup $\langle a^4\rangle$ contains 
$$\{a^4,a^8,a^{12},a^{18},a^{20},a^{24},a^{28},a^{32}=a^2,a^6,a^{10},a^{14},a^{18},a^{22},a^{26},a^{30}=e\}.$$
So by Lagrange theorem the index $[\langle a\rangle:\langle a^4\rangle]=2$.
