# Why is a multiplicative inverse preserved in a group homomorphism but lost in a ring homomorphism?

Take $$\varphi: R_1 \longrightarrow R_2$$, and $$f:G_1 \longrightarrow G_2$$.

In order to prove $$\varphi$$ is a ring homomorphism, we must show $$\varphi(1_{R_1})=1_{R_2}$$ (in addition to other properties), but to prove $$f$$ is a group homomorphism it suffices just to show $$f(g_1\cdot g_2) = f(g_1)\cdot f(g_2)$$, with $$f(e_{G_1})=e_{G_2}$$ being implied - how is this so?

Referencing this post about why we need to specify $$\varphi(1_{R_1})=1_{R_2}$$, the zero map is brought up as an example of why we need this specification, since $$0_{R2}$$ might not be the identity element of $$R_2$$, but $$\varphi(1_{R_1}\cdot r) = \varphi(1_{R_1})\cdot\varphi(r)= 0 \cdot 0$$, so $$0_{R_2} \textbf{will}$$ be the identity element of $$\varphi(R_1)$$.

But doesn't this same reasoning apply to our group homomorphism $$f$$? The following proof from my textbook for why $$f(e_{G_1}) = e_{G_2}$$ seems like it should theoretically hold for rings...

$$e\cdot g = g$$

$$f(e\cdot g) = f(g)$$

$$\Rightarrow f(e) \cdot f(g) = f(g)$$

$$\Rightarrow f(e)$$ is the identity element of $$G_2$$

• You can't cancel by $f(g)$ in a ring. Feb 28, 2022 at 3:49
• It sounds like you're mixing up the category of rings [with multiplicative identity] and the category of rngs [which may not have a multiplicative identity]. There's a theory for both, and in the category of rngs, when $R_1$ has a multiplicative identity $1_{R_1}$, its image $f(1_{R_1}) \in R_2$ will always be the multiplicative identity of the rng $f(R_1)$ even when $R_2$ doesn't have a multiplicative identity. This is, in essence, what you seem to be saying. Feb 28, 2022 at 4:26

The last implication assumes that $$f(g)$$ is invertible (since you only multiplied by $$f(g)^{-1}$$ to the right on both sides to get that $$f(e)$$ is the identity element), and this might not be the case for a ring, since there are ring elements that are not invertible. In the counterexample that you pointed out, as you said, $$f$$ could be the zero map and, in this case, $$f(x)$$ is not invertible, so the argument used for group homomorphism does not apply there.