Mathematics Asked by Tota on January 25, 2021
How to prove that there is not a monomorphism from Klein 4-group to $Z_6$(or a epimorphism from $Z_6$ to $V_4$) ?
I know that:
If $f$ is a monomorphism than $Kernel(f)={0}$ .
In $V_4$ , three elements have order 2.
In $Z_6$ , $’3’$ has order 2, $’2’$ and $’4’$ have order 3 , $’1’$ and $’5’$ have order 6.
I read a theorem which says:"If $G$ and $G’$ are finite groups and $f$ is a homomorphism between them so that $f(a)=a’$ , where $a$ is from $G$ and $a’$ is from $G’$ , then the order of $a’$ divides the order of $a$."
My question is:Using this theorem, can I say that the monomorphism between $V_4$ and $Z_6$ does not exist since orders $3$ and $5$ do not divide $2$(order of all elements of Klein group)?
Alternatively:
$Z_6$ is cyclic and so all its subgroups are cyclic. Since $V_4$ is not cyclic, it cannot be embedded into $Z_6$.
Every homomorphic image of a cyclic group is cyclic. Since $V_4$ is not cyclic, it cannot be an homomorphic image of $Z_6$.
Correct answer by lhf on January 25, 2021
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