Some hours ago I asked a question about condition(s) for a generator to be mapped to a generator by a homomorphism, but I think I was not making myself clear what I was asking for, and sometimes even confused myself. So this time I want to make the following question more productive and useful.
suppose $\phi:\Bbb Z/m\Bbb Z \rightarrow \Bbb Z/n\Bbb Z$ is a homomorphism.
1.If $\phi$ is bijective, must $\phi$ map a generator to a generator? (Yup)
2.If $\phi$ is injective but not surjective, must $\phi$ map a generator to a generator? (Nope)
3.If $\phi$ is surjective but not injective, must $\phi$ map a generator to a generator? (????)
1.If so, we write $\phi:\Bbb Z/m\Bbb Z \rightarrow \Bbb Z/n\Bbb Z$, where $m=n$. We show that if $x$ generates the cyclic group, then$|x|=|\phi(x)|$. Since $\phi$ is a homomorphism,
-$\phi(x)^{|x|}=\phi(x^{|x|})=\phi(1)=1$, the identity
-$x^{|\phi(x)|}=\phi^{-1}(\phi(x)^{|\phi(x)|})=\phi^{-1}(\phi(1))=1$, the identity
We can do this by defining $\phi^{-1}:\phi(x)\rightarrow x$. This is well-defined, since $\phi(x)=\phi(y)$ implies $x=y$ ($\phi$ is injective). One can easily check that $\phi^{-1}$ is a homomorphism.
but I noticed that $\phi^{-1}$ needs not be defined on all of $\Bbb Z/n\Bbb Z$, the codomain of $\phi$. It just has to be defined on the image of $\Bbb Z/m\Bbb Z$ under $\phi$.
2. No. we write $\phi:\Bbb Z/m\Bbb Z \rightarrow \Bbb Z/n\Bbb Z$, where $m<n$. If $x$ is a generator of $\Bbb Z/m\Bbb Z$, then clearly $\phi(x)$ cant generate all of $\Bbb Z/n\Bbb Z$. or else let $\phi(x)^k$ be arbitrary element of $\Bbb Z/n\Bbb Z$, $k$ an integer. Then $\phi(x)^k=\phi(x^k)$, contradicts to the fact that $\phi$ is not surjective.
3. Lets take an example: let $x_{36}$ be the usual generator of $\Bbb Z/36\Bbb Z$, $x_{4}$ be that of $\Bbb Z/4\Bbb Z$
$\phi:\Bbb Z/36\Bbb Z \rightarrow \Bbb Z/4\Bbb Z$ defined by $\phi(x_{36})=x_4$. This map is well-defined, since if $x_{36}^a=x_{36}^b$, then $36|{a-b}$, and so $4|{a-b}$, $x_4^a=x_4^b$. It certainly is a surjective homomorphism.
$x_{36}, x_{36}^5, x_{36}^7,x_{36}^{11},x_{36}^{13},x_{36}^{17},x_{36}^{19},x_{36}^{23},x_{36}^{25},x_{36}^{29},x_{36}^{31},x_{36}^{35}$are all the generators of $\Bbb Z/36\Bbb Z$, and they're all mapped to either $x_4$ or $x_4^3$, the generators of $\Bbb Z/4\Bbb Z$ by $\phi$.
But if we now take another example: define $\phi:\Bbb Z/36\Bbb Z \rightarrow \Bbb Z/35\Bbb Z$ by $\phi(x_{36})=x_{35}^t$. For $\phi$ to be well-defined, $x_{36}^a=x_{36}^b$ must imply $({x_{35}^t})^a=({x_{35}^t})^b$, i.e. $36|{a-b}$ must imply $35|t(a-b)$. Since $36$ and $35$ are coprime, $35|t$. i.e. $\phi$ is the trivial homomorphism. It certainly does not map a generator to a generator.
Now write $\phi:\Bbb Z/36\Bbb Z \rightarrow \Bbb Z/8\Bbb Z$ defined by $\phi(x_{36})=x_{8}^t$. This time, for $\phi$ to be well defined, $2|t$, but $\phi(x_{36})=x_8^{2t}$, certainly not a generator of $\Bbb Z/8\Bbb Z$.
Questions:
-Can someone help me generalize the results of 3.?
-In the proof of 1., the fact that $\phi$ is injective was made use. It seems that $\phi$ needs not be surjective, but in 2., if $\phi$ is injective but not surjective, $\phi$ does not map a generator to a generator. What's wrong here?
For (1), it suffices to show (3).
For (2), it suffices to find a counterexample. Consider the embedding (injective homomorphism) $\phi: \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$ via $x \mapsto 2x$.
For (3), yes. Let $\phi: \mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ be a surjective homomorphism. Suppose $g$ generates $\mathbb{Z}/m\mathbb{Z}$. Then $\mathbb{Z}/m\mathbb{Z} = \{e, g, g^2, \dots g^{m - 1}\}$. By surjectivity of $\phi$, $\mathbb{Z}/n\mathbb{Z} = \phi(\mathbb{Z}/m\mathbb{Z}) = \{e, \phi(g), \phi(g)^2, \dots \phi(g)^{m - 1}\}$. So $\phi(g)$ generates $\mathbb{Z}/n\mathbb{Z}$.