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I have to find non trivial homomorphisms $\varphi$, $\varphi'$: $\mathbb{Z}/3\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/7\mathbb{Z})$. We know that Aut $(\mathbb{Z}/7\mathbb{Z}) \cong (\mathbb{Z}/7\mathbb{Z})^\times \cong \mathbb{Z}/6\mathbb{Z}$. So we have to look order of elements of $\mathbb{Z}/6\mathbb{Z}$ that divide $3$. These elements are $[2]_6$ and $[3]_6$. In corrections these homomorphisms are defined as: $\Big(\varphi([1]_3)\Big)([i]_7) = [2i]_7$ and $\varphi'$:$\Big(\varphi([1]_3)\Big)([i]_7) = [4i]_7$. The thing that i don't understand, is what $[i]_7$ represents. For me, if we generalize for example $\varphi$, we got $\varphi([r]_3) = [2^r]_7$. I read a lot of articles on ''research'' of homomorphisms, and i feel confused right now and can't get an intuition for this. Thanks in advance for help.

Shaun
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Daniil
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    $\varphi([1]_3)$ is an automorphism of $\Bbb{Z/7Z}$, $\varphi([2]_3)=\varphi([1]_3)\circ \varphi([1]_3), \varphi([3]_3)=Id$ and $[i]_7,[2i]_7$ are elements of $\Bbb{Z/7Z}$. – reuns Dec 16 '20 at 12:45
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    There is no need to invest so much work on notations, thus also making notations full with "white noise" (parantheses and indices). It is enough to write $(\Bbb Z/3, +)\to (\Bbb Z/6, +)\cong ((\Bbb Z/7)^\times, \cdot)\to\operatorname{Aut}(\Bbb Z/7, +)$. (Which is also an exagerated pedant way to write things.) Then mention the image of the generator $1\in\Bbb Z/3$ via the steps used for $\phi$. Something like $1\to 2\to 3^2\to(3^2\cdot)$ defines $\phi$. (Here, $3$ is my choice of a multiplicative generator in the ring/field $\Bbb Z/7$, we are implicitly using the ring structure.) – dan_fulea Dec 16 '20 at 12:48
  • See also this similar post with the comments. – Dietrich Burde Dec 16 '20 at 12:53
  • @dan_fulea Thank you very much for your explanation, but i still can't understant how you get $1 \rightarrow 2 \rightarrow 3^2\rightarrow(3^2⋅)$. If you could elaborate a bit more, i would really appreciate it, Thank you! – Daniil Dec 16 '20 at 13:16
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    I am just defining $\phi$ as follows. A map from one group $G$ to an other one $H$ is determined by the image of the generators of $G$. (Of couse, there are constraints... but it is determined.) So we need to fix the image of $1=[1]_3\in \Bbb Z/3=G$ via $\phi$ as an element of $H$, which is Aut of $(\Bbb Z/7,+)$. The image is obtained by moving $1$ from group to group in the above chain. In your notations, $[1]_3\to[2]_6\to[3]_7^2\to($multiplication by $[3]_7^2)$. Simply written, $1\to 2\to 3^2\to(3^2\cdot)$. (For $r\in G$ it's $\phi:r\to 2r\to 3^{2r}\to(3^{2r}\cdot)$ and $\phi':r\to4r\to$...) – dan_fulea Dec 16 '20 at 13:35
  • @dan_fulea Oh i think i got the logic. My problem was that once i determined a morphism from $\mathbb{Z}/3\mathbb{Z} \rightarrow \mathbb{Z}/6\mathbb{Z}$, i thought that it is equivalent to construct a morphism $\mathbb{Z}/3\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/7\mathbb{Z})$ directly ''on basis'' of first that i have just written. But now i understand that you have to pass from $\mathbb{Z}/6\mathbb{Z}$ by $(\mathbb{Z}/7\mathbb{Z})^\times$ to get in $\text{Aut}(\mathbb{Z}/7\mathbb{Z})$. If its false let me know please! Anyway, thank you very much! – Daniil Dec 16 '20 at 13:47
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    I think you got it right. (It is sometimes important when working with "algebraic structures" to have a "categorial view", so having the focus on objects (here groups (with supplementary structure)) and on maps between them. The elementwise approach is often not the good way to view things. It is sometimes the only way to describe a map, but if a map can be "broken into pieces" (structurally), it is a good idea to do that.) Please always ask, do not hesitate to do it. (You've got a downvote, i also get'em, don't let this discourage your questions, just get better in presenting your problems.) – dan_fulea Dec 16 '20 at 14:10

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