# When is it true that $\nexists g\ne e$ such that $\phi(g)=e$ for a group homomorphism $\phi: G\mapsto H$ and $|H|\not\mid|G|$?

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 $$\phi(g)=e$$?

• 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. – lulu Jan 4 at 13:18
• 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 – John Cataldo Jan 4 at 13:20
• 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 $H$. – lulu Jan 4 at 13:22
• 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 $G$. – Omnomnomnom Jan 4 at 13:22
• And indeed, if $|G| \nmid |H|$, then there are no injective homomorphisms from $G$ to $H$. – Omnomnomnom Jan 4 at 13:25

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.

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