I am struggling with a lemma allegedly from a paper of Eilenberg and Moore from back in the nascent days of category theory. I encountered it in Rotman's Group Theory text as an exercise and couldn't track it down anywhere else, from which I will reproduce the statement:
(Lemma) Let $G$ be a group, with subgroup $1 < H < G$. Then there exists a group $L$ and distinct homomorphisms $f, g: G \rightarrow L$ s.t. $f|H \neq g|H$.
So I just don't understand what this theorem is saying, really. If we let $L = G \times G$, then we can just let $f, g$ be the component embeddings. Or we could let $L = G$ and take $f, g$ to be the two trivial homomorphisms.
That is too easy so I seriously doubt Eilenberg and Moore had that in mind. So what I'm assuming is that the term 'distinct' means the following:
(Lemma*) Let $G$ be a group, with subgroup $1 < H < G$. Then there exists a group $L$ and a unique pair of homomorphisms $f, g \in$ Hom$(G, L)$ s.t. $f|H \neq g|H$.
Now the lemma is really weird. If $L, f, g$ are as proscribed, then we have $(f \circ \sigma)|H = f|H$ for all $\sigma \in $Aut$(G)$ (and the same for $g$), but being able to fuse up to $H$ along the entirety of Aut$(G)$ seems impossible in a typical scenario.
So, what is the correct interpretation of this exercise? Possibly helpful is the fact that the author gives a hint, which tells you to consider an augmented coset action: Let $X = \lbrace aH$ $|$ $a \in G\rbrace$ $\cup$ $\lbrace \infty \rbrace$ and let $G$ act on $X$ in the normal manner, where $a \cdot \infty = \infty$. It forms a proper action clearly, so we can let $f$ be the embedding of $G$ into $S_X = L$ induced from this action. Consider $\gamma \in S_X$, where $\gamma$ is conjugation by $(H, \infty)$, the transposition exchanging $H$ and $\infty$. He says to let $g = \gamma \circ f$, which can also be seen to be a homomorphism easily.
So the implication, then, is that any other homomorphism from $G \rightarrow L$ is equal to one of these two when restricted to $H$. I don't see why that's true. Of interest is the fact that $f$ collapses precisely $H$ to the identity, which should imply that $H$ is normal. I just have no idea what the point of this exercise is, it seems wrong in more ways than one etc.
Any help?
I'm pretty sure that it's a typo and should read $f|H = g|H$.
Then the claim is equivalent to the fact that epimorphisms in the category of groups are surjective, which I believe was indeed proved by Eilenberg and Moore in the "nascent days of category theory".