Homomorphism $f: G \to H$ maps $e_g$ to $e_h$

I'm trying to wok out the proof that a homomorphism, $f: G \to H$ maps the identity in one group to the identity in another. Several other answers provide an explanation for this, but there is one step I am confused on and haven't been able to find by browsing past answers. It's likely that I'm misunderstanding the justification for drawing on the inverse of $f$ and the logic which follows.

starting with the standard property of homorphisms, $f(g_1 * g_2) = f(g_1) \oplus f(g_2)$, taking $*$ to be the binary operation in $G$, $\oplus$ to the binary operation in $H$, and $g_1, g_2 \in G$.

From here, the proof lets $g_1 = g_2 = e_g$, where $e_g$ is the identity of the group, $G$, which implies that $f(e_g * e_g) = f(e_g) * f(e_g)$, so $f(e_g) = f(e_g) * f(e_g)$.

Then, we multiply on the left by $\left(f(e_g)\right)^{-1}$, which should cause this to simplfiy to the identity on the left and $f(e_g)$ on the right. Here is my confusion:

(a) How do we conclude that such an inverse exists? I believe that a homorphism is by definition bijective and, by extension, invertible, as establishing such a function is necessary to establish that two groups are isomorphic. Is this the argument? Or is there an additional condition specific to the particular groups?

(b) Which binary operation are we applying when we multiply by this inverse? Clearly, $f(e_g) \in H$ by the definition of $f$, which has the operation $\oplus$, but $\left(f(e_g)\right)^{-1} \in G$, which has the operation, $*$. How can we 'mix' operations in this sense between elements of two different groups?

Thanks.

• The inverse exists because $H$ is a group. You're applying the operation of $H$. Commented Jul 15, 2018 at 0:58
• You also moved from calling the operation of $H$ $\oplus$ to $*$ about halfway through, which created your confusion in question (b). Commented Jul 15, 2018 at 0:59

Here's what it should look like: $$f(e_G) = f(e_G * e_G) = f(e_G) \oplus f(e_G).$$ Ignore the middle term and write $$e_H \oplus f(e_G) = f(e_G) \oplus f(e_G).$$ Note that all I have done is replace $f(e_G)$ by $e_H \oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^{-1}$ computed in $H$.

• Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^{-1} \oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– user465188
Commented Jul 15, 2018 at 1:16
• My equation is in $H$, so everything is in $H$. "On the right" means $\text{blah blah blah} \oplus [f(e_G)]^{-1}$. Do that to both sides. Commented Jul 15, 2018 at 1:22

How do we conclude that such an inverse exists?

Since $H$ is a group every element has its unique inverse.

I believe that a homorphism is by definition bijective

This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense. But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.

ad b): You just use the property, that you have a homomorphism.