Proving that a group homomorphism preserves the identity element

Question

Assume that $(G,*)$ and $(H,o)$ are groups and that $f:(G,*) \rightarrow (H,o)$ is a homomorphism.

Let $e_G$ and $e_H$ denote the identity elements of $G$ and $H$, respectively. Show that $f(e_G)=e_H$.

Approach: $f(e_G)=f(a*a^{-1})$ for $a,a^{-1} \in G$, so $f(a*a^{-1})=f(a)of(a^{-1})$.

If that’s true, then how do we know that $f(a^{-1})$ is the inverse of $f(a)$?

Answer 1

Using the fact that $e_G\ast e_G=e_G$ yields $$ f(e_G)=f(e_G\ast e_G)=f(e_G)\circ f(e_G)$$

Now multiply both sides by the inverse of $f(e_G)$ to obtain $f(e_G)=e_H$.

Answer 2

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So if $a\in G$ $$f(a) = f(a* e_G) = f(a)\circ f(e_G).$$

Left cancel by $f(a)^{-1}$ to get $e_H$

Answer 3

We have $e_G = e_G \ast e_G$, then

$$f(e_G) = f(e_g \ast e_G) = f(e_G)\circ f(e_G).$$

Now, $$e_H = [f(e_G)]^{-1}\circ f(e_G) = [f(e_G)]^{-1}\circ (f(e_G)f(e_G)) = f(e_G).$$