Check whether a map is an automorphism

Question

Let $G$ be a cyclic group of order $12$. Check whether a map $T:G\ni x \mapsto x^3\in G$ is an automorphism of $G$.

My solution looks like an overkill. By Cauchy's Theorem there exists $x \in G, x \ne e$ such that the order of $x$ is $3$. Assume that the given map is an automorphism of $G$. Therefore, $T(x)=x^3=e=T(e)$. By the injectivity of $T$ we must have $x=e$ which is absurd. It follows that $T$ is not an automorphism of $G$.

I'm looking for a more elementary solution with may using the property that $G$ is cyclic. Any help would be much appreciated.

Answer 1

I might be missing something, but it looks like it's not injective:

Suppose $G=\langle g \rangle$.

Then $T(g)=g^3$, and $T(g^5)=(g^5)^3=g^{15}=g^3$, but $g\neq g^5$.

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About your solution:

You don't need Cauchy's theorem to know there is an element of order $3$. The order of $x^k$ is $n/d$, where $n$ is the order of $G$ and $d=\gcd(k,n)$; so, if $G=\langle g\rangle$, an element of order $3$ is $g^4$. (And another one would be $g^8$.)

Answer 2

We can as well let $G$ to be $\mathbb Z_{12}$. It is easy to see that $x\to 3x$ is not an injection since $3.8=3.4$ but $8\ne 4$.