# Finding subgroups if an element has an order

Question 1: If $a$ has order $12$, find the subgroups $\langle a^{3}\rangle$ and $\langle a^{5}\rangle$. Also is $\langle a^{2}\rangle$ a subgroup of $\langle a^{4}\rangle$ ?
Question 1b:If $\langle b^{r}\rangle$ is a subgroup of $\langle b^{s}\rangle$, how are $r$ and $s$ related?

Here is my attempt on question 1:
$\langle a^{3}\rangle =\langle e, a^{3},a^{6},a^{9} \rangle$

$\langle a^{5}\rangle=\langle e,a^{5},a^{10},a^{3},a^{8},a,a^{6},a^{11},a^{4},a^{9},a^{2},a^{7} \rangle$ note: I did not put them order because my professor said not to so he can follow along when he check it. Basically, I am doing it as I go along

Here is the other part to question 1:
$\langle a^{2} \rangle$=$\langle e,a^{4},a^{6},a^{8},a^{10} \rangle$
$\langle a^{4} \rangle$=$\langle e,a^{4},a^{8} \rangle$
Is $\langle a^{2}\rangle$ a subgroup of $\langle a^{4}\rangle$? I said No

Here is my attempt to question 1b:
$r$ and $s$ are related because of their order. I am not sure how to correctly answer this question though.

If somebody can check this for me and see where did I go wrong with this problem please let me know.If I am wrong correct me if you can

• Question 1 looks fine, once you point out that $\langle a^5\rangle$ actually is the same as $\langle a\rangle$, and give a concrete reason for why $\langle a^2\rangle$ is not a subgroup of $\langle a^4\rangle$ (which is only a matter of saying that $a^2\notin \langle a^4\rangle$). – Arthur Apr 16 '17 at 18:50
• For question 1b, suppose $H=\langle b^r \rangle \leq \langle b^s \rangle = K$, then for every $z \in \mathbb{Z}$, $b^{rz} \in K$, so there exists $x \in \mathbb{Z}$ such that $b^{rz}=b^{sx}$. Note that in general $rz$ might not be equal to $sx$, this will depend on the group containing $H$ and $K$ (think about $\mathbb{Z}$ and $\mathbb{Z}_n$). – Gilberto López Apr 16 '17 at 19:50