Question: Find $Aut\left ( \mathbb{Z}_{6} \right )$
Note that $\mathbb{Z}_{6}=\left \{ 0,1,2,3,4,5 \right \}$
Observe:
$\forall k \in \mathbb{Z}_{6}$, $k^{6}=e \equiv 0(mod6)$
Recall: Suppose $\phi$ is an isomorphism from a group G to G. Then $\forall a \in G \left | \left ( a \right )\phi \right |=\left | a \right |$
But $\alpha$ is an isomorphism so is by definition also a Homomorphism. So, $\forall k \in \mathbb{Z}_{6} \left ( k \right )\alpha=\left ( 1+\cdot \cdot \cdot +1\right )\alpha=\left ( 1 \right )\alpha+\cdot \cdot \cdot +\left ( 1 \right )\alpha=k\left ( 1 \right )\alpha$
Recall: $Aut\left ( \mathbb{Z}_{n} \right )\cong U\left ( n \right )$
So, $Aut\left ( \mathbb{Z}_{6} \right )\cong U(6)=\left \{ 1,5 \right \}$
Thus, we deduce the possible candidates to be $\left ( 1 \right )\alpha=1$ or $\left ( 1 \right )\alpha=5$
Here is where I am unable to progress further. Any help is appreciated.
$\endgroup$ 12 Answers
$\begingroup$Your approach starts off in the right direction, but then suddenly
Recall $\operatorname{Aut}(\Bbb{Z}_n)\cong U(n)$
So $\operatorname{Aut}(\Bbb{Z}_6)\cong U(6)=\{1,5\}$
Then what's left to prove? And why bother with the rest of your argument?
$\endgroup$ 2 $\begingroup$To answer the question, it is enough to show that $Aut(C_6)$ has $\phi(6)=2$ elements. This was proved here, for example. Then we have $$ Aut(C_6)\simeq C_2. $$ Furthermore it was shown at the duplicate, that $$ Aut(C_n)\simeq U(n)=(\mathbb{Z}/n\mathbb{Z})^{\ast}. $$
$\endgroup$
