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Let $G=\langle a \rangle$, $o(G)=12$ be a cyclic group, then $T:G \to G$ s.t. $T(x)=x^3$ is not an automorphism.

I know $T$ is not an automorphism, and I have seen proofs where $T$ is not 1-1. But here is my attempt with another approach:

Let $G=\langle a \rangle$. $T$ is not an automorphism of $G$ since there is no $x \in G$ s.t. $T(x)=a^5$. That is, if $x = a^i$ for some integer $i$, then $T(x) = a^{3i}$. So $a^{3i}=a^5$, but there is no integer $i$ for this to be possible. So $T$ is not onto, hence, not an automorphism.

Is my argument correct?

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    $\begingroup$ How do you know $a^{3i}\ne a^5$? Note: in $\mathbb Z/(11)$ under $+$, $9\times3\equiv5$ $\endgroup$ – J. W. Tanner Mar 8 '20 at 2:40
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Yes. But you should prove it. You must show $3i+12k=5$ has no solution. This is a linear diophantine equation. It's true there are no solutions, since $\operatorname{gcd}(12,3)=3\nmid 5$.

Here's an alternate proof, using a basic fact about cyclic groups: $|a^3|=12/\operatorname{gcd}(3,12)=4$. Thus the image only has order $4$, and the map is not surjective.

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For a cyclic group of order $n$, the $k$th power map is an automorphism if and only if $\gcd(k,n)=1$. In particular, since $3$ and $12$ are not relatively prime, cubing on a cyclic group of order $12$ is not an automorphism.

Either showing the failure of injectivity or of surjectivity is sufficient to conclude that an endomorphism is not an automorphism. For example, $a^4$ is a non-identity element that cubes to the identity, showing that injectivity fails. On the other hand, $a$ itself is not a cube, showing that surjectivity also fails.

For endomorphisms of finite groups, injectivity and surjectivity are actually equivalent to each other. Possibly infinite groups for which surjective endomorphisms are also injective are called "Hopfian groups", while those for which injective endomorphisms are also surjective are called "co-Hopfian groups". In particular, a cyclic group of order $12$, like any other finite group, is both Hopfian and co-Hopfian.

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