How to prove for a finite group that $|\langle g \rangle|=o(g)$?  I really don't know how to show that this is true I tried to say let $o(g)=m$ and then show that $|\langle g \rangle|$ has exactly $m$ elements but I wasn't even sure it did because it would contain the elements $g^0,...g^{m-1}$ so that would mean it's true but what about $g^{-1}$ and the other elements with negative coefficients so it seems like it would actually be bigger than $m$?  Thanks.
Please note: I posted this question about an hour ago on my account (I just made it as I am new to the site). I revisited my post and saw that I had not given definitions of what I was asking (I thought they were common notations) and it would not let me edit the post for some reason so I will ask here now and hopefully someone can delete the other thread.  How to prove for a finite group $G$ that $|\langle g \rangle|=o(g)$?  Here are the definitions:  Let $G$ be a finite group and $g \in G$  $o(g)$ is the smallest $m \in \Bbb{N}$ such that $g^m=e$ (the identity element of the group). (If such an $m$ exists)  $\langle g \rangle=\{g^m : m \in \mathbb{Z}\}$  I really don't know how to show that this is true I tried to say let $o(g)=m$ and then show that $|\langle g \rangle|$ has exactly $m$ elements but I wasn't even sure it did because it would contain the elements $g^0,...g^{m-1}$ so that would mean it's true but what about $g^{-1}$ and the other elements with negative coefficients so it seems like it would actually be bigger than $m$?  Thanks.
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