TITLE: Does there exist a group homomorphism $f:\Bbb{Z}/6\Bbb{Z} \to \Bbb{Z}/12\Bbb{Z} $ with $f(\bar 1)=\bar 3$
QUESTION [0 upvotes]: Does there exist a group homomorphism 
$f:\Bbb{Z}/6\Bbb{Z} \to \Bbb{Z}/12\Bbb{Z} $
$f(\bar 1)=\bar 3$
So what i have so far is
$f(\bar 0)=0$
$f(\bar 1)=3$
$f(\bar 2)=6$
$f(\bar 3)=9$
so $|Img(f)|=4$
but i dont know how to continue, can i please get some tips. Maybe first isomorphism theorem helps, or is there a shorter way ?

REPLY [2 votes]: Suppose $f(1)=3$.
Note that $0=6$ in $\Bbb{Z}_6$.
Hence $f(0)=f(6)$.
Since $f$ is a homomorphism, $f(0)=0$.
Also, $f(6)=6f(1)=6\cdot3=18$.
Hence we get $0=18$ in $\Bbb{Z}_{12}$ which is a contradiction.

REPLY [1 votes]: Why did you stop at $f(3) = 9$? What happens next, and is that allowed?
What I mean is, $f(4) = 12 = 0$, and $f(5) = 3$, and $f(0) = f(6) = 6$, and this is a problem!
In general, if $o(x) = \min\{ n > 0 : nx = 0\}$ is the order of $x$ (the group being written additively) and if $f$ is a group homomorphism, then $o(f(x))$ divides $o(x)$. In particular, $f(1)$ cannot equal $3$ because the order of $3$ in $\mathbf{Z}/12$ is $4$ whereas the order of $1$ in $\mathbf{Z}/6$ is $6$, and $4$ does not divide $6$.

REPLY [0 votes]: $f([1])=[3]$ ;$\Bbb Z_6=\langle [1]\rangle \implies f[x]=f(x.[1])=x.f([1])=[3x]$
Now it remains to check whether $f([x+y])=[3(x+y)]=[3x]+[3y]=f([x])+f([y])$
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