It it always true that the image of the neutral element in a group homomorphism $f$ is the neutral element of the codomain group regardless whether $f$ is injective/surjective?

The answer is most probably true as written at various places but I can’t understand the following (counter?) example.

Let $G$ be the group $\{0,1\}$ and the mapping $f: G \rightarrow G$ such that for all $x \in G, f(x) = 0$ ; $f$ is indeed a group morphism $f(x*y) = f(x)*f(y)=0$ but the image of the neutral $f(1)$ is $0$ and different to $1$.


  • $\begingroup$ Are you considering the two elements $0$ and $1$ with their usual multiplication? That's not a group. $0$ has no inverse. $\endgroup$ – badjohn Feb 11 at 8:33

As I said in a comment, your $G$ does not seem to be a group, what is the inverse of $0$?

Let's suppose $f : G \rightarrow H$ is a group homomorphism.

The identity of $G$ is $e_G$ and that of $H$ is $e_H$.

Let's distinguish two cases:

  1. $\forall g \in G: f(g) = e_H$

Obviously, $f$ maps $e_G$ to $e_H$.

  1. $\exists g \in G: f(g) \neq e_H$

Let $h = f(g)$.

$f(e_Gg) = f(e_G)f(g)$ since $f$ is a homomorphism.

But also $e_G g = g$ so:

$f(g) = f(e_G)f(g)$

$h = f(e_G) h$

$H$ is a group so $h$ has an inverse $h^{-1}$.

$h h^{-1} = f(e_G) h h^{-1}$

$e_H = f(e_G)$.

So the image of $e_G$ is $e_H$.

We did not actually need to distinguish the two cases, it just seemed a little clearer to do so.

  • $\begingroup$ Thank you. Very clear. $\endgroup$ – mauricebis Feb 11 at 9:29

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