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I was thinking if it is possible to have $g\ne e$ such that $\phi(g)=e$ for a group homomorphism $\phi: G\mapsto H$

It's not always true because $\phi(2)=0$ when $G=\Bbb Z_4$ and $H=\Bbb Z_2$

So what if $|H|\not\mid|G|$?

We have that $o(\phi(g))$ divides $o(g)$ and $|H|$ because $\phi(\langle g\rangle)$ is a subgroup of $H$

If we had $(|H|,\phi(\langle g\rangle))=1$ then we would have $o(\phi(g))=1$ and so $\phi(g)=e$

What other necessary/sufficient constraints are there for the existance of $g\in G\backslash\{e\}$ with $\phi(g)=e$?

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    $\begingroup$ Not sure I get the question. For any groups $G,H$ you always have the trivial homomorphism $G\to H$ which takes every element to the identity. $\endgroup$ – lulu Jan 4 at 13:18
  • $\begingroup$ My question is about the conditions that would assure $\exists g\in G~~:~\phi(g\ne e)=e$. So $\phi=id$ is a condition but my question is more general $\endgroup$ – John Cataldo Jan 4 at 13:20
  • $\begingroup$ Well, if $\phi$ has no non-trivial elements in its kernel then $\phi(G)$ is isomorphic to $G$, in which case of course we have $|H|\, |\,|G|$ (assuming the groups to be finite, which you never state). Is that what you wanted? Conversely, if $G$ is isomorphic to a subgroup of $H$ then that isomorphism gives you an injection from $G$ to $H$. $\endgroup$ – lulu Jan 4 at 13:22
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    $\begingroup$ A more concise way to rephrase your question (as I understand it): under what conditions are there no injective homomorphisms from $G$ to $H$. The answer to this is that an injective homomorphism exists iff $H$ contains a subgroup isomorphic to $G$. $\endgroup$ – Omnomnomnom Jan 4 at 13:22
  • $\begingroup$ And indeed, if $|G| \nmid |H|$, then there are no injective homomorphisms from $G$ to $H$. $\endgroup$ – Omnomnomnom Jan 4 at 13:25
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First of all, let's use a fact to make your question a bit neater.

The following conditions are equivalent:

  • $\phi : G \to H$ is injective (one-to-one)
  • $\ker \phi = \{e\}$
  • There does not exist an element $g \in G$ with $g \neq e$ such that $\phi(g) = e$

With this in mind, I believe that you are asking this: under what conditions does there exist an injective homomorphism from $G$ to $H$? The answer to this question is that such a homomorphism will exist if and only if $H$ contains a subgroup isomorphic to $G$. For the example of $G = \Bbb Z_2$ and $H = \Bbb Z_4$, we see that the map $\phi([n]_2) = [2n]_4$ is injective, and $\Bbb Z_4$ contains the subgroup $\{[0]_4,[2]_4\}$ which is isomorphic to $\Bbb Z_2$.

One way to prove this is to use the first isomorphism theorem. In particular, we know that for any homomorphism, $\phi(G) \cong G/\ker\phi$. However, if $\ker \phi = \{e\}$, then we have $G/\ker \phi \cong G$. So, if $\phi$ is an injective homomorphism, then $\phi(G) \cong G$, and $\phi(G)$ (the image of $\phi$) is a subgroup of $H$.

Conversely, suppose that $H$ has a subgroup $K \subset H$ and that $K \cong G$. Then, an isomorphism $\phi:G \to K$ means that we have the injective homomorphism $i_K \circ \phi:G \to H$, where $i_K:K \to H$ is the inclusion map.

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  • $\begingroup$ @Mark good catch, thanks $\endgroup$ – Omnomnomnom Jan 4 at 13:42
  • $\begingroup$ Thanks! (I think you meant big $G$ and $K\cong G$) $\endgroup$ – John Cataldo Jan 4 at 13:49
  • $\begingroup$ @JohnCataldo I don't see what you mean by "big $G$" $\endgroup$ – Omnomnomnom Jan 4 at 13:50
  • $\begingroup$ "$H$ contains a subgroup isomorphic to $g$" $\endgroup$ – John Cataldo Jan 4 at 13:51
  • $\begingroup$ @JohnCataldo that makes sense now, thanks $\endgroup$ – Omnomnomnom Jan 4 at 13:53

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