Are these two equivalent interpretations of "The only homomorphic images of $G$ are $1$ and $G$"? Let $G$ be a group, say nontrivial.  Aluffi interprets the property 
"The only homomorphic images of $G$ are $1$ and $G$ [paraphrased]" ($\dagger$)
to mean
"If there exists a surjective homomorphism $\varphi \colon G \twoheadrightarrow G'$, then either $G' \cong 1$ or $G' \cong G$ [paraphrased]." $(*)$
The context is we want to show such groups $G$ satisfying $(*)$ are simple.  The way to do this is to assume otherwise, take a nontrivial proper normal subgroup $N$ of $G$, and look at $\pi \colon G \twoheadrightarrow G/N$.  
Question:  How can we show it is impossible that $G \cong G/N$ under some other homomorphism $\varphi$?  For example, we could have something like $\prod^\infty \mathbb{Z} \cong \prod^\infty \mathbb{Z} / (\mathbb{Z}, 0, 0, \dots)$.  Of course, this group doesn't satisfy $(*)$ (just project onto the first coordinate), so it isn't exactly a counterexample, but could some nasty group exist that is? What prevents this?  
Is it better to think of $(\dagger)$ to mean
"Every surjective homomorphism out of $G$ is either trivial or an isomorphism?" $(**)$
Is this what Aluffi actually meant by $(*)$?  Does it not matter?  That is, are $(*)$ and $(**)$ actually equivalent as written?
Thanks in advance.
Related:  Definition of Simple Group
 A: The first statement is ambiguous. One way to read "$G' \cong G$" is that it asks for an isomorphism as abstract groups, and this gives something which is not obviously equivalent to the second statement; as you say, there could be some strange group $G$ with the property that all of its quotients are trivial or abstractly isomorphic to $G$ even if some of those quotients are nontrivial. I don't know an example off the top of my head though. 
Another way, which is equivalent to the second statement, is to require that the isomorphism $G' \cong G$ is not only an isomorphism of groups but an isomorphism of quotient groups; here, if $\varphi_1 : G \to G_1$ and $\varphi_2 : G \to G_2$ are two quotient groups of $G$ (the quotient maps $\varphi_1, \varphi_2$ are part of the data), then an isomorphism of quotient groups between $(G_1, \varphi_1)$ and $(G_2, \varphi_2)$ is an isomorphism $f : G_1 \to G_2$ which in addition satisfies $f \circ \varphi_1 = \varphi_2$. 
A: The selected answer to the post linked by user1729 gives an example of a non-simple group satisfying $(*)$, and so this answers my question, with the conclusion that $(*)$ does not necessarily imply $(**)$ (of course, $(**)$ implies $(*)$!).  
