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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!

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up vote 6 down vote accepted

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?

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i dont get why the generator g generates upto g^(p-2) shudnt it be p-1?? im confused! – d13 Dec 11 '12 at 8:28
What happens when you start squaring elements? You will need that $g^{p-1} = e$. – JavaMan Dec 11 '12 at 8:40
To more directly answer your last question, note that $G$ is of order $p-1$ so that $g^{p-1} = e$, where $e$ is the identity. – JavaMan Dec 11 '12 at 8:52
I'm not sure what you mean when you say "I have to use group homomorphism." All you need to notice is that $H = \{e, g^2, g^4, \dots , g^{p-3}\}$. This set has cardinality $\frac{p-1}{2}$. – JavaMan Dec 11 '12 at 9:04

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