# Given $\phi\colon G\rightarrow G'$ onto, does $\phi(G)=G'$?

Working through Fraleigh's "Abstract Algebra ($$7$$th ed)" and came across lemma $$8.15$$ which says:

Given groups $$G$$, $$G'$$ and an isomorphism $$\phi\colon G\to G'$$, then $$\phi(G)$$ (the image of G under $$\phi$$) is a subgroup of $$G'$$.

My question is, why doesn't $$\phi(G)=G'$$? Surely since the isomorphism is onto by definition, we must have the image of the domain is the whole codomain? Is it just a matter of saying that $$\phi(G)$$ is actually equal to $$G'$$, but since $$G'$$ is a trivial subgroup of itself it just makes it easier to state it like this?

• I don't have my copy available, but are you sure that the book said $\phi$ is an isomorphism and not a homomorphism? Commented Apr 8, 2020 at 11:01
• Yes you are totally right, an isomorphism is onto so the image is the total of $G'$. Sometimes in the literature it is meant that $G$ embeds isomorphically into $G'$, so then an injective homomorphism $\phi$ would be enough and then $G \cong \phi(G)$ of course. Commented Apr 8, 2020 at 11:02
• In my copy (seventh edition) this is not the statement. The statement I have is: "Let $G$ and $G'$ be groups and let $\phi:G \to G'$ be a one-to-one function such that $\phi (xy) = \phi (x) \phi (y)$ for all $x,y \in G$. Then $\phi [G]$ is a subgroup of $G'$ and $\phi$ provides an isomorphism of $G$ with $\phi[G]$"
– user649348
Commented Apr 8, 2020 at 11:08
• $\phi(G)$ is a subgroup of $G'$, even when the group homomorphism $\phi$ is not bijective, injective or onto. Commented Apr 8, 2020 at 11:10
• Aye, looking back at the textbook and actually reading it properly I've realised this is what it says. Commented Apr 8, 2020 at 11:11

You are completely right regarding your question stated in the title: if $$\phi\colon G\to G'$$ is onto (i.e. surjective) then by definition the image $$\phi(G)$$ equals all of the codomain, that is $$\phi(G)=G'$$.

Anyway, I have to agree with André Armatowski. In my copy Lemma $$8.15$$ also reads as

Lemma $$\textbf{8.15}$$ Let $$G$$ and $$G'$$ be groups and let $$\phi\colon G\to G'$$ be a one-to-one function such that $$\phi(xy)=\phi(x)\phi(y)$$ for all $$x,y\in G$$. Then $$\phi[G]$$ is a subgroup of $$G'$$ and $$\phi$$ provides an isomorphism.

First off, the requirement that $$\phi$$ is one-to-one is unnecessary for the conclusion that $$\phi[G]$$ (I will adopt Fraleigh's notation) is a subgroup. To see this, consider a group homomorphism $$\varphi\colon G\to H$$ that is, a set-function $$\varphi\colon G\to H$$ such that $$\varphi(xy)=\varphi(x)\varphi(y)~\forall x,y\in G$$.

Claim. The image $$\varphi[G]$$ of $$G$$ under $$\varphi$$ is a subgroup of $$H$$.

Proof. As one can show directly from $$\varphi$$ being a group homomorphism, we have that $$\varphi(e_G)=e_H$$ and that $$\varphi(g^{-1})=\varphi(g)^{-1}$$. Therefore $$\varphi[G]$$ is non-empty. Now, let $$\tilde x,\tilde y\in\varphi[G]$$ then there exist $$x,y\in G$$ such that $$\varphi(x)=\tilde x$$ and $$\varphi(y)=\tilde y$$, respectively. Thus $$\tilde x\tilde y^{-1}=\varphi(x)\varphi(y)^{-1}=\varphi(x)\varphi(y^{-1})=\varphi(xy^{-1})\in\varphi[G]$$ By the subgroup test, we conclude that $$\varphi[G]$$ is a subgroup of $$H$$.$$~~~\square$$

Lemma $$8.15$$ evolves around how we can obtain a trivial isomorphism from an injective function $$\phi$$ by restricting the codomain to the image of $$\phi$$. By definition, $$\phi$$ is surjective on its image and as $$\phi$$ is injective by assumption we obtain an isomorphism, i.e. a bijective group homomorphism. That's it.

• @m1820 If anything remains unclear, please let me know so that I can adjust my answer accordingly :) Commented Apr 9, 2020 at 16:15
• @m1820 Anything I can add to make you consider accepting the answer (and consequently closing up this question thread)? Commented Oct 25, 2021 at 0:57