# Is this proof that a homomorphism preserves identities correct (sufficient)? [closed]

My textbook writes the following proof for the statement in the title:

For a homomorphism $$\varphi:G \to G'$$ and any $$g\in G$$, $$\varphi(g) = \varphi(ge) = \varphi(g)\varphi(e)$$. Multiplying both sides of this equation on the left by $$(\varphi(g))^{−1}$$ gives the desired result, $$e′ = e′\varphi(e) = \varphi(e)$$.

It's not clear to me why we should be able to work like this. In particular, we only have that $$\varphi(g) = \varphi(g)\varphi(e)$$ for all elements in $$G'$$ of the form $$\varphi(g)$$. But it seems like the proof is tacitly using that $$e'$$ is the unique element for which $$e'g' = g'$$ for all $$g \in G'$$...and yet we haven't shown that $$\varphi(e)$$ actually does this for every $$g'$$!

• No, this is not tacitly assumed. We just take the equation $\phi(g)=\phi(g)\phi(e)$ and cancel the invertible element $\phi(g)$. But $\phi(g)\cdot \phi(g)^{-1}=1$ in $G'$, and the $1$ is the $e'$. We obtain $e'=1=\phi(e)$. Take for example $G=G'=\Bbb Z$ and $\phi(n)=2n$. Mar 19 at 20:21
• Use $\varphi$ for $\varphi$ and $\in$ for $\in$. Mar 19 at 20:23
• @DietrichBurde I guess I am looking for an argument which doesn't just say "cancel" but shows what that means. For example, suppose we have the lemma that in a group $G$, if $gh = gk$ then $h=k$.. Computing, we have $h = eh = (g^{-1}g)h = g^{-1}(gh) = g^{-1}(gk) = (g^{-1}g) k = ek = k.$ I was trying to convince myself using this more direct line.
– EE18
Mar 19 at 20:25
• Cancelling in a group really means $g\cdot g^{-1}=e$, so $gh=gh'$ gives $h=h'$. This is all. So it is just the existence of $g^{-1}$. The cancelling in the multiplicative group of a field, say, of real numbers is the same. If $2a=2b$, then $a=b$, because $2$ has an inverse $1/2$, which we can multiply from the left. Mar 19 at 20:26
• In a group, if $xy=x$ for a single $x$, then $y=e$. Mar 19 at 20:29

Because every element in a group has an inverse, it's enough for

$$fg' = g'$$

for just one $$g'\in G'$$ to prove that $$f$$ is the identity.

The fact that we already know that $$G'$$ is a group has done most of the work for us - we know that there is an element in $$G'$$ that preserves every element when you multiply by it, we just have to show that that's what $$\varphi$$ maps $$e$$ to.

• Ah I see. I think I can use the lemma I proved in the comments above to show this?
– EE18
Mar 19 at 20:27
• Yup, that lemma is a great proof. And it's exactly what mathematicians mean when they say "cancel" in a group setting. :-) Mar 19 at 20:31
• Awesome, thank you!
– EE18
Mar 19 at 20:38
• As an aside, does this mean that if we relax the "for any $g \in G$" and talk about a map such that for exactly one $g \in G$ we have $f(g) = f(ge) = f(g)f(e)$ then $f$ preserves identities?
– EE18
Mar 19 at 20:57
• Hmm, I don't think I understand what you are asking., especially with the "exactly one $g$" specification. $f(g) = f(ge)$ automatically for all $g \in G$, not just one. - that's $e$ being the identity in $G$. And $f(ge) = f(g)f(e)$ is also true for all $g$ - that's what being a homomorphism is. If you're asking if we can have weaker assumptions than having a group identity or the homomorphism rule, and still deduce preserving identities, then maybe you should check out monoids or semi-groups and see how things that work in groups can stop working in those structures. Mar 19 at 21:24