# Proving that a group homomorphism preserves the identity element

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 $f(a)$?

• You don't, that would require $f(e)=e$. But try your same argument but with $a=e$ and it will work, we know $e^{-1}=e$ right. Aug 1, 2016 at 4:13
• I think you can carry on with $f(a)f(a^{-1}) = f(a)f(a)^{-1} = e_H$. If you are allowed to use $f(a^{-1}) = f(a)^{-1}$ Aug 1, 2016 at 4:26
• @Nameless, that's what I have to prove next Aug 1, 2016 at 4:26
• @TheMathNoob, well I guess you can't use what I suggested. You need help with that one or...? Aug 1, 2016 at 4:27
• @TheMathNoob, well actually we can more or less use what did here. Since you know $f(a)f(a^{-1}) = e_H$, what must $f(a^{-1})$ be? Actually I think my initial comment was backwards. Aug 1, 2016 at 4:30

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 $f(e_G)=e_H$.

• ok so $f(e_G)=f(e_G) \circ f(e_G)$, $f(e_G) \circ f^{-1}(e_G)=(f(e_G)\circ f(e_G)) \circ f^{-1}(e_G)$ and hence $\circ$ is associative we can say $f(e_G)=e_H$ Aug 1, 2016 at 4:22
• That's correct. Aug 1, 2016 at 4:23
• That trick seemed to be very useless XD. Aug 1, 2016 at 4:24

The answers provided here are excellent, but here is something to just widen your mind for diversity.

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 $e_H$

• Oh woahhh XDDDDD Aug 1, 2016 at 4:37

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