I'd like to see how $Aut(Z_{49})\cong Z_{42}$. I know that $Aut(Z_{49})\cong (\mathbb{Z}/42\mathbb{Z})^x$ where $Z_k$ is the cyclic group order k, but I fail to see how either of these is isomorphic to $Z_{42}$, or even that it is cyclic for that matter.
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$(\mathbb{Z}/42\mathbb{Z})$ is the multiplicative group by the way – xiaolung tintin yao Nov 18 '15 at 07:48
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Check that $3$ is of order $42$ in $Z_{49}^*$... – Clément Guérin Nov 18 '15 at 08:06
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is this part of a more general theory where id be able to arrive at the same conclusion had 49 been replaced by n – xiaolung tintin yao Nov 18 '15 at 08:07
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You have that $\mathrm{Aut}(\mathbb{Z}_{49}) \simeq \mathbb{Z}_{49}^{\times}$, not $\mathbb{Z}_{42}^{\times}$. Now you want to show that $\mathbb{Z}_{49}^{\times} \simeq \mathbb{Z}_{42}$.
This group does not have too many elements, so just check to find a generator. Once you get more than 21 distinct powers of an element, you are guaranteed that element is a generator.
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