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Let $G$ and $H$ be groups, and suppose I want to construct a group homomorphism $\phi$ between them. From what I know, I just need to send each element $x \in G$ to an element $y \in H$ such that the order of $\phi(x)=y$ divides the order of $x$. Is that right?

I was a bit confused, because I was trying to construct a nontrivial homomorphism from $D_{6} = \{1, b, b^2, a, ba, b^2a\}$ to $\Bbb{Z}_2 = \{e, s\}$ , and I thought of sending $a \in D_6$ to $s \in \Bbb{Z}_2$, since the order of $s$ (obviously) divides that of $a$, and sending the rest of the elements to $e$. However, we have then have $\phi(ba)\phi(b) = (e)(e) = e$ and $\phi(bab) = \phi((ab^2)b)) = \phi(a) = s \not=e$. So $\phi$ cannot be a group homomorphism, right?

Am I doing something wrong here, or do I need to satisfy more conditions in order to construct a group homomorphism?

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Your example is a good counterexample to the false claim "a map $\phi : G \to H$ that satisfies $\operatorname{order}_H(\phi(x))$ divides $\operatorname{order}_G(x)$ for all $x \in G$ is a homomorphism." It is true that any homomorphism satisfies this condition (i.e., it is a necessary condition) but it is not sufficient, as your example shows.

A homomorphism simply must satisfy $\phi(x)\phi(y) = \phi(xy)$ for any $x,y \in G$.

Here are some hints for constructing your nontrivial homomorphism:

  • The set of elements of $G$ that map to $e$ is a subgroup of $G$ known as the kernel of $\phi$.
  • So, if $x$ and $y$ are not in the kernel of $\phi$, then $xy$ must be in the kernel because $\phi(x)\phi(y)=s^2=e$.
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No, your condition is necessary but not sufficient. e.g. any map from the Klein 4-group $\langle a,b:a^2=b^2=aba^{-1}b^{-1}=e\rangle$ to itself that fixes the identity satisfies your condition, but any group homomorphism is determined by where it sends $a$ and $b$.

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To construct a homomorphism between (G,*) and (H,•)

For every pair of elements g1, g2 in G we have: f(g1 * g2) = f(g1) • f(g2)

for this to be true, you need to map the generetor in G to the generator in H and keep on.

For example; constructing a non trivial homoomorphism between Z/6Z and Z/3Z both additive:

map 1+ Z/6Z to go to 1 in Z/3Z, map 2+ Z/6Z to go to 2 in Z/3Z, map 3+ Z/6Z to go to 0 in Z/3Z, map 4+ Z/6Z to go to 1 in Z/3Z, map 5+ Z/6Z to go to 2 in Z/3Z, map 0+ Z/6Z to go to 0 in Z/3Z

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