0
$\begingroup$

Let $G$ and $H$ be groups and $p \in \mathbb{N}$ be a prime such that $p|o(G)$ and $p|o(H)$. Then, we know that there exist $g$ and $h$ in $G$ and $H$ respectively which generate cyclic subgroups of order $p$. Then, given an arbitrary $x \in G$, $x=yg^m$ for some $y \in G$ and $ 0 \leq m < p $. Define $$ \phi : G \to H,$$ $$ \phi(yg^m)=h^m.$$ Then, this map is a homomorphism so far as $\phi$ is well-defined. This basically boils down to whether different representations of $x \in G$ affect the image of $x$ under $\phi$. I don't know how to proceed if I want to prove or disprove the well-definedness of this map.

$\endgroup$
1
  • 1
    $\begingroup$ $g=g\cdot g^0=g^0\cdot g^1$. What should $\phi(g)$ be? $\endgroup$
    – awllower
    Feb 25 '20 at 14:32
3
$\begingroup$

Your $\phi$ is never well-defined, the representation $x=yg^m$ is never unique or valid. Your "some $y\in G$" is simply $xg^{-m}$ and can be taken for any $m$. Thus, by your definition, we have $\phi(x)=h^m$ for any $m$, which makes sense only when $h$ is the neutral element.

In fact two groups having orders with a non-trivial common divisor is not enough for a non-trivial homomorphism to exist. Even when the orders are equal. Consider any non-abelian simple group $G$ (e.g. the alternating group $A_n$) and any abelian group $H$ such that $|H|=|G|$, e.g. $H=\mathbb{Z}_{|G|}$. Note that no non-trivial homomorphism $G\to H$ exists.

This is true, however, if $G$ is assumed to be abelian, or more generally when $G$ has a normal subgroup of index $p$. That's because in this situation there's a quotient map $G\to\mathbb{Z}_p$ which composed with an embedding $\mathbb{Z}_p\to H$ gives a non-trivial homomorphism.

$\endgroup$
0
$\begingroup$

Here, checking that the map $\phi$ is well defined amounts to the following task :

Task
Check if $h^m$ depends on the choice of $y, m$ such that $x = yg^m$.

In other words, you have to check whether it is possible to write $x = y g^m$ and $x = y'g^{m'}$ with $m \neq m'$ or not. If it is possible, $\phi$ isn't well defined. If it isn't possible, $\phi$ is well defined.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.