Index of $\langle a^4\rangle$ in the group $\langle a\rangle$

Question

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?

Answer 1

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 $a \langle a^4 \rangle$ ).

Answer 2

Hint: Let $o(a)$ denotes the order of $a$, then for any positive integer $n$ we have $$o(a^n)=\frac{o(a)}{\gcd(o(a),n)}.$$

Answer 3

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 $[\langle a\rangle:\langle a^4\rangle]=2$.

Answer 4

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!!!