The order of a subgroup of a cyclic subgroup

Question

Let $G = Z^*_p$ under multiplication, with $p$ being a prime $> 3$ and $|G|=p-1$ and $G$ is cyclic. $H = \{a^2\mid a \in G\}$. Want to prove $H < G$ and $|H| = (p-1)/2$.

I can show that $H < G$, but how do I find the order of $H$ if I have to use the natural group homomorphism?

I know that $|H|$ divides $p-1$.

Thank you for your help!

Answer 1

Large-ish Hint:

First, note that $|H| \leq \frac{p-1}{2}$ as $H \neq G$ coupled with your observation that $|H|$ divides $|G|$.

Since $G$ is cyclic, we can write $G = \langle g \rangle = \{e, g, g^2, \dots , g^{p-2}\}$. Finally, consider $H = \{g^{2k} \mid 0 \leq k \leq p-2\}$. What do you notice?