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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 $\forall a \in G, \left |\phi \left ( a \right ) \right |=\left | a \right |$

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.

AKP2002
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Mathematicing
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2 Answers2

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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?

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Servaes
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  • My mind wasn't thinking well while working on these problems on board a plane. The real question that has been bothering me is this: $\alpha$ is an isomorphism from a group to itself. It is thus self-implied that $\alpha$ is a bijection and so each element is the image of a distinct element in the domain. In this sense, how is it possible for $(1)\alpha=1$ and $(1)\alpha=5$? For if this were true then the bijection map would be false. – Mathematicing Jul 01 '16 at 10:04
  • You've shown that there are two automorphisms of $\Bbb{Z}_6$, determined by mapping $1\in\Bbb{Z}_6$ to either $1$ or $5$. – Servaes Jul 01 '16 at 12:00
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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 $$ Aut(C_n)\simeq U(n)=(\mathbb{Z}/n\mathbb{Z})^{\ast}. $$

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Dietrich Burde
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