Studying groups of order $12$, specifically, the semi-direct products that appear in these groups, I come across something that I do not fully understand.
In Dummit (pag. 183 groups order 12) you need to see the amount of homomorphisms that there between $T\simeq \mathbb{Z}/{3\mathbb{Z}}$ and ${\rm Aut}(V)\simeq S_3$. The argument they give is the following:
$\mathbb{Z}/{3\mathbb{Z}}$ is cyclic. Then let $T=\langle y\rangle $and there is a unique subgroup of ${\rm Aut}(V)$ of order $3$, say $\langle \gamma\rangle$ . This, there are three possible homomorphisms from $T$ into ${\rm Aut}(V)$:
$\varphi_i:T\to{\rm Aut}(V)$ defined by $\varphi_i(y)=\gamma^i$, $i=0,1,2.$
Question: Why are there only three possible homomorphisms from $T$ into ${\rm Aut}(V)$?
I know it has to do with the order of the domain of the homomorphism but I don't see why the possible homomorphisms need to reach a subgroup with the same order of their domain (for the homomorphism to be well defined)
Best Answer
Say $G$ is cyclic of order $n$ with generator $g$ and $T \colon G \to K$ a group morphism. Then $T(g)^n = T(g^n) = T(1) = 1$, so the order of $T(g)$ divides $n$. Likewise, if $x \in K$ has order divisible by $n$, you can check that $T(g^k) := x^k$ yields a morphism.
Hence $\hom_{\mathsf{Grp}}(\mathbb{Z}/n\mathbb{Z}, K)$ is in correspondence with the elements of $K$ whose order is divisible by $n$. When $n=3$ there are not many options: these are precisely the elements of order $1$ or $3$.
Now, an element $x$ of order $3$ generates a subgroup $\langle x\rangle = \{1,x,x^2\}$ of order three. If there is only one such group in $K$, then all elements of order three are just $x$ and $x^2$. Together with $1$, they determine all maps $\mathbb{Z}/3\mathbb{Z} \to K$.