Is the following mappings automorphisms of their respective groups? (Herstein BOOK)

Question

Is the following map an automorphism of the cyclic group $G$ of order $12$ ? $$T : x \rightarrow x^3 $$

My attempt :

I thinks the answer is yes, because $G$ is cyclic, $$T(xy) = (xy)^ 3= x^3y^3 =T(x)T(y)$$ So $T$ is a homomorphism

Also $T$ is one -one as $T(x) = T(y) $ implies $x=y$

Is my logic correct or not ?

Any hints/solution will be appreciated

Thanks

Answer 1

Yes, $T$ is a homomorphism, but that's because $G$ is Abelian, which implies that $(xy)^3=x^3y^3$. And, no, it is not an automorphism. If $G=\langle a\rangle$, then $(a^4)^3=(a^8)^3$, but $a^4\neq a^8$.

Answer 2

It isn't an automorphism. $\mathbb{Z}/12\mathbb{Z}$ is a cyclic group of order 12. With your transformation you get $T(0)=0$ and $T(4)=0$ so it's not one to one.

Answer 3

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.