Slight generalization of the First Isomorphism Theorem

Question

I'm being asked to prove the following generalization of the First Isomorphism Theorem:

If $f:G \mapsto H$ is a homomorphism, and $N $ is a normal subgroup of $G$ contained in $ker(f)$, with $\pi$ as the quotient homomorphism, that there exists a unique homomorphism $f':G/N \mapsto H$ with $f' \pi =f$

As far as I can tell this is almost identical to the first isomorphism theorem with the only major difference being that $N$ is not equal to the kernel. At first glance, this makes me think that this may disrupt $f'$ being well defined, but since proving that it is depends on all $n$ in $N$ having the property that $f(n)=e$ (the identity element in G), but if $N$ is contained in the kernel, that would take care of itself.

Is there something else I'm missing? I find it unlikely that I'm just being asked to spew back a proof we've already seen in class?

I will also add, I understand that if $f$ is surjective we will see $f'$ mapping to $H$ and if $f$ is not surjective, we'll be mapping to $Img(f)$ instead, and I'm not sure if this is a strange way of asking about that generalization.

Thanks in advance!

Answer 1

This result is quite similar to the first isomorphism theorem, but with some minor subtleties, which are not so obvious (to me, at least) and can be instructive. What follows is a sketch of the proof.

Let $\bar{f}: G/\ker(f) \to H$ be the unique morphism such that $\bar{f}\pi_{\ker(f)} = f$, given by the first isomorphism theorem. Given $N \subseteq \ker(f), N \triangleleft G$, we always have a morphism

$$ \tau : G/N \longrightarrow G/\ker(f) \\ gN \mapsto g\cdot\ker(f) $$

Now, we can take $f' = \bar{f}\tau$, since for every $g \in G$,

$$ f'\pi(g) = f'(gN) = \bar{f}\tau(gN) = \bar{f}(g\ker(f)) = \bar{f}\pi_{\ker(f)}(g) = f(g) $$

and the uniqueness is given by the following fact: since $\pi$ is an epimorphism, it is right cancellative, that is $g\pi = h\pi$ implies $g = h$. In particular if we have another morphism $f'':G/N \to H$ such that $f''\pi = f$, then

$$ f''\pi = f = f'\pi $$

and so $f'' = f'$.

As you said, we're very strongly using that $N \subseteq \ker(f)$, but nevertheless there are still some remarks to be made.

I've also noted that you talked about $f'$ being surjective or not, sometimes the first isomorphism theorem is stated for surjective morphisms, so that we get $G/\ker(f) \simeq H$, but more generally, if you have a morphism $\phi:G \to H$, by the first isomorphism theorem we have a morphism $\phi': G/\ker(f) \to \operatorname{im}(f)$ with $\phi'\pi = \phi$, by taking $\iota\phi'$ with $\iota:im(f) \to H$ the inclusion map, you now have a (non-necessarily surjective) morphism $\iota\phi':G \to H$, which is unique (we can again invoke that $\pi$ is right cancellative) and satisfies $\iota\phi'\pi = \phi$.

Answer 2

This proof is going to be an introduction to diagram chasing. I suggest you try it yourself first, it was really fun to do. I'll list the full proof out below (doesn't require reproving first iso theorem, but uses it quite a bit). The question that's really being asked is why do we get a unique homomorphism $f':G/N\to H$.

I suggest you draw out a diagram of the maps/groups I'm about to say. Your diagram should have G,H,G/N and G/K for the groups, where $K$ denotes $\ker(f)$. I'll list out the maps along the way of the proof, although you could probably guess half the maps off the bat.

The basic idea is to factor $\pi_K:G\to G/K$ into $\pi_K=\star \circ \pi_N$ where $\star:G/N\to G/K$ denotes some map I'll define in a second. We know that by first iso, there exists an unique $\tilde{f}$ such that $f=\tilde{f}\circ \pi_K$. You're going to define $f'=\tilde{f}\circ \star$. I claim that $\star$ is well defined (as a group homomorphism) and unique. This is because we'll define $\pi_K=\star \circ \pi_N$, and by naturality of $\pi$, since this is really the only way we can define $\star$, so we have uniquness. Thus, $f'$ exists and is unique because $\tilde{f}$ and $\star$ were individually.

It remains to show $\star$ is well defined and $\pi_K=\star\circ \pi_N$. Take $\star:\bar{g}\mapsto gK$, and $$\star(\bar{g_1})\star(\bar{g_2})=g_1Kg_2K=g_1g_2K=\star{(\overline{g_1g_2})=\star(\bar{g_1}\cdot \bar{g_2})}$$ remember bar is a homomorphism, thus the notation works out. Next, for any $g\in G$, $$\pi_K(g)=gK=\star{(\bar{g})}=(\star\circ\pi_N)(g)$$