Let $G=\langle a\rangle $ cyclic group of order $9$ If $H=\langle a^2\rangle $ prove that $H=G$ Let $G=\langle a\rangle $ cyclic group of order $9$
H subgroup of G
If $H=\langle a^2\rangle $ prove that $H=G$
Is the order of H 2? Then since H subgroup of G $\Rightarrow$ H cyclic.
then since 2 is prime, $H=\{e\}$  or $H=G$
Im really confused  what am I missing?
 A: I think that you are missing the fact that, if the order of $H$ was really $2$, then, since $G$ has order $9$, there would be no way for you to have $H=G$.
Note that$$\overbrace{(a^2)^5}^{\phantom H\in H}=a^{10}=a^9\cdot a=a,$$and that therefore $a\in H$. But $G=\langle a\rangle$. So, $H=G$.
A: A more general useful result is the following:

Theorem: Let $G$ be a group and $g \in G$ with $o(g) = n <\infty$. Then $o(g^m) = \frac{n}{\gcd(n, m)}$.

This is a relatively well-known fact but if you have never encountered it you can find its proof here.
Applying this to your specific problem, $|G| = o(a) = 9$, $H \leq G$ and $|H|=o(a^2) = \frac{9}{\gcd(9, 2)} = 9$. From that we conclude that $G=H$.
A: Here is another way to prove it

Theorem For any Finite group $(G,*)$, if $a\in G$, $\langle a\rangle=\langle a^k\rangle$ if $\gcd(\text{ord}(a),k)=1$.

($G$ no need to be cyclic)
Proof
Let $(G,*)$ be a finite group and $a\in G$, then $\gcd(k,\text{ord}(a))=1$ implies $\text{ord}(a)x+ky=1$ for some $x,y\in\mathbb{Z}$. Therefore
$$a^{\text{ord}(a)x+ky}=a^{\text{ord}(a)x}*a^{ky}=a$$
$$\implies a^{ky}=a$$
$$\implies \langle a^{k}\rangle=\langle a \rangle $$
A: You can think of your $G$ as $(\Bbb Z_9,+)$ and, accordingly, $\langle a^2\rangle$ as $\langle 2\rangle$. You can prove explicitly that $\langle 2\rangle$ spans the whole $\Bbb Z_9$, by multiplying (additive notation!) $2$ by one, two, three, etc. up to nine (modulo $9$).
