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The automorphism group of the group $G$, $\text{Aut}(G)$, is the group of isomorphisms from $G$ to $G$.

It is the set of functions $\phi$ that send the generators of $G$ to the generators of $G$, right?

I want to show that $\text{Aut}(\mathbb{Z}_6)$ is isomorphic to $\mathbb{Z}_2$.

Since the generators of $\mathbb{Z}_6$ are $1$ and $5$, $\text{Aut}(\mathbb{Z}_6)$ is the set of functions that send $1$ to $1$ or to $5$ and $5$ to $1$ or $5$, or not?

Such a function is the identity function and the function $x\mapsto 5x$, right?

Which function do we use to show the isomorphism?

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  • $\begingroup$ In your second line: it in fact is the set of bijective homomorphisms, not only functions from $\;G\;$ to itself. $\endgroup$ – DonAntonio Feb 17 '16 at 19:39
  • $\begingroup$ Which other functions besides from $G$ ? @Joanpemo $\endgroup$ – Mary Star Feb 17 '16 at 19:40
  • $\begingroup$ All the permutations on $\;G\;$ as a set, @Mary Star. Most of these ones aren't automorphisms (group theory), just bijections (set theory). $\endgroup$ – DonAntonio Feb 17 '16 at 19:42
  • $\begingroup$ Ah ok... I am a little confused how to find a function to show the isomorphism... Could you give me a hint? @Joanpemo $\endgroup$ – Mary Star Feb 17 '16 at 19:46
  • $\begingroup$ You don't need to explicitly define such a fuction. You have already answered the question. Remember that the group of automorphisms is by no means finite. They can be classes of mappings. $\endgroup$ – Phillip Hamilton Feb 17 '16 at 19:49
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I think you're in the right track: any homomorphism from any group $\;G\;$ towards any group is completely determined by its action on some generating set of $\;G\;$. If we talk of automorphisms then generators are mapped to generators

In our case, $\;G:=\Bbb Z_6\;$ is a cyclic group of order 6, and thus an automorphism is completely determined by its action on a generator of $\;G\;$.

Since there are two elements of order six here, namely $\;1,5\pmod 6\;$ , every automorphism must map any of them into itself or the other one, and thus clearly Aut$\,(G)\cong\Bbb Z_2\;$

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    $\begingroup$ So every automorphism group of a group that has $2$ generators is isomorphic to $\mathbb{Z}_2$ because each generator is either mapped to itself or to the other elemens, as it happens in $\mathbb{Z}_2$. Have I understood it correctly? $\endgroup$ – Mary Star Feb 17 '16 at 19:54
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    $\begingroup$ @MaryStar Yes...but some care is due here. When you say "a group that has two generators" one can understand a group that NEEDS two elements to be generated. That is not our case here. What we have here is that $\;G\;$ , being cyclic, can be generated by one single element, and there are two possible candidates in this case to be that single element generating the group. For example $\;S_3\;$ is a group that needs at least two elements to be generated by them. $\endgroup$ – DonAntonio Feb 17 '16 at 19:56
  • $\begingroup$ Ah ok... I see... Is this justification enough or do we have to proof the isomorphism by defining a function and showing that it is an homomorphism and a bijection? $\endgroup$ – Mary Star Feb 17 '16 at 20:53
  • $\begingroup$ We have that the possible generators of $\mathbb{Z}_8$ are $1,5$. The automorphisms map one generator to one of the others. So, the automorphisms are the following functions: $$f_1(x)=x, f_2(x)=5x$$ Could we show that the two groups are isomorphic to each other by defining the function $$h:\text{Aut}(\mathbb{Z}_6)\rightarrow \mathbb{Z}_2 \text{ with } \\ f_1\mapsto 0\pmod 2 \\ f_2\mapsto 1\pmod 2$$ and show that it is bijective and an homomorphism? $\endgroup$ – Mary Star Feb 18 '16 at 18:28
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    $\begingroup$ I think you meant $\;\Bbb Z_6\;$ above, and yes: your $\;h\;$ works fine, in my opinion. $\endgroup$ – DonAntonio Feb 18 '16 at 18:39

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