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I know that $$(\ \mathbb{Z} /49 \mathbb{Z} )\ ^* \simeq \mathbb{Z}/ 42 \mathbb{Z} $$ because $\phi(49) = 42$. Furthermore $$\mathbb{Z}/ 42 \mathbb{Z} \simeq \mathbb{Z}/ 6 \mathbb{Z} \times \mathbb{Z}/ 7 \mathbb{Z}$$

Is the first claiming always true? The cardinality of the two sets is equal but isomorphism is a stronger fact. So... why does this thing work?

The second one I know is a consequence of the chinese theorem remainder, so it is true when $\mathbb{Z}/ n\mathbb{Z}$ factorises with coprime factors.

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No, the group $(\mathbb{Z}/n\mathbb{Z})^{\ast}$ is not always cyclic. It is the case for prime powers $p^n$, two times prime powers $2p^n$ for $p>2$, and $n=1,2,4$. Have look here:

When is the group of units in $\mathbb{Z}_n$ cyclic?

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  • $\begingroup$ So when is cyclic what I said is true? $\endgroup$ – user515933 Jan 28 '18 at 22:10
  • $\begingroup$ A cyclic group of order $n$ is always isomorphic to $C_n$, so then it is always true. Otherwise it need not be true. Take $n=4$. Then $C_4$ and $C_2\times C_2$ are two non-isomorphic groups of order $4$. $\endgroup$ – Dietrich Burde Jan 28 '18 at 22:11

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