Understanding short exact sequences I'm trying to understand the presentation on short exact sequences in Benedict Gross's algebra lecture, but I'm having difficulty. This is the definition he gives (paraphrased):

Given $G,H,G'$ groups, consider
$$1 \to H \to G \to G' \to 1,$$
and let $g: H \to G$ and $f: G \to G'$ be homomorphisms, where $g$ is one-to-one, $f$ is onto, and $g(H) = $ the kernel of $f$.

Defining the maps in this way is fine. If $g$ is one-to-one, it is injective so it's kernel is trivial. The statement $g(H) = \mathrm{ker(f)}$ doesn't make complete sense to me. If $g$ is injective its kernel is trivial, and $g$ has to map identity to identity because it's a homomorphism, so it has to map the kernel in $g$ into the kernel in $H$. If $g$ is injective, does not that not imply that $H$ is the trivial subgroup?
Another reading of this (based on some lecture notes) states instead $\mathrm{Im}(g) = \mathrm{ker}(f)$. Is this the correct notion? Is there an interplay between injectivity of $f$, surjectivity of $f$, and this fact?
Finally, he concludes that by $G \cong G/H$ by the first isomorphism theorem. This would require that $H$ be a normal subgroup of $G$ (though it's possible this was embedded in the assumptions). $f: G \to G'$ is surjective by assumption, so we have $G/\mathrm{ker}(f) \cong \mathrm{im}(f)$ by the first isomorphism theorem. So this would require that $H = \mathrm{ker}(f)$ (I suppose this is where the above assumption comes into play).
I'm also a little bit confused about the "$1$'s": this is strange notation to me, and I'd be inclined to write $\{e\}$. But are they different identities? Is this the identity in $H$, $G$, or $G'$? Does it not matter, since the groups are equivalent up to isomorphism anyway?
I would greatly appreciate any help on parsing this concept.
 A: A few things you said show a misunderstanding of the situation:

The statement $g(H) = \mathrm{ker(f)}$ doesn't make complete sense to me. If $g$ is injective its kernel is trivial, and $g$ has to map identity to identity because it's a homomorphism, so it has to map the kernel in $g$ into the kernel in $H$. If $g$ is injective, does not that not imply that $H$ is the trivial subgroup?

It is correct that the kernel of $g$ is trivial and $g(e_H)=e_G$. However, it is unclear what you mean by "the kernel in $g$" and "the kernel in $H$". A group doesn't come with a kernel, a homomorphism does. In this situation, as you said the kernel of $g$ is trivial, $\ker(g)=\{e_H\}$, and the kernel of $f$ is assumed to be the image of $g$, so $\ker(f)=g(H)=\operatorname{im}(g)$. Note that $g$ being injective makes it an isomorphism onto its image, so $H\cong\ker(f)$ via $h\mapsto g(h)$. So no, $H$ neither needs to be trivial nor does it need to be a subgroup of $G$. $H$ needs to be a group isomorphic to the kernel of $f$.

Another reading of this (based on some lecture notes) states instead $\mathrm{Im}(g) = \mathrm{ker}(f)$. Is this the correct notion? Is there an interplay between injectivity of $f$, surjectivity of $f$, and this fact?

Since $\operatorname{im}(g)$ or $\operatorname{Im}(g)$ is just another notation for $g(H)$, both conditions are equivalent and hence both are correct.
There is indeed an interplay of the three conditions. We already figured out that injectivity of $g$ together with $\operatorname{im}(g)=\ker(f)$ yields $H\cong \ker(f)$. Now surjectivity of $f$ tells us that $f$ induces an isomorphism $G/\ker(f) \cong G'$ given by $[g]\mapsto f(g)$. Putting things together we may write
$$
G/g(H) \cong G',\quad\text{where $g(H)\cong H$}.
$$
Some authors are a bit sloppy in this situation and write $G/H$ instead of $G/g(H)$. I prefer to always make clear that the quotient is with respect to $g(H)$, a subgroup of $G$ isomorphic to $H$ via $g$. The group $H$ need not even be a subgroup of $G$ so $G/H$ is an abuse of notation. Also note that the isomorphism type of a quotient $G/U$ is not determined by the isomorphism types of $G$ and $U$: a group $G$ can have normal subgroups $U_1,U_2$ with $U_1\cong U_2$ but $G/U_1\ncong G/U_2$. This makes the abuse of notation in "$G/H$" worse, since we really need to know how $H$ is embedded into $G$.
Since kernels are always normal subgroups and $g(H)=\ker(f)$, we indeed always have that $g(H)$ is a normal subgroup of $G$.

I'm also a little bit confused about the "$1$'s": this is strange notation to me, and I'd be inclined to write $\{e\}$. But are they different identities? Is this the identity in $H$, $G$, or $G'$? Does it not matter, since the groups are equivalent up to isomorphism anyway?

The notation "$1$" here is just a usual convention to write the group with only one element, which is unique up to unique isomorphism. It is none of "the identity in $H$, $G$ or $G'$", since those are elements of a group and here $1$ denotes a group itself. You might write $1=\{e_1\}$ to be precise about the names of the identity elements in all groups. However, most authors will not make this distinction and use "$1$" to denote both the group with one element and any identity element of a (multiplicatively written) group.
A: In general, an exact complex just means a sequence of groups
$$\dots \to  G_{i-1} \to G_i \to G_{i+1} \to \dots$$
where $f_i : G_{i-1} \to G_i$ and $\ker f_i = \operatorname{Im}(f_{i-1})$.
In your above example, saying $g(H) = \ker f$ is saying that if $x \in G$ and $f(x) = 0$ then there is some $y \in H$ such that $x = g(y)$. The kernel of $f$ is precisely the image of $g$.
It might be more helpful to think about it as a diagram
$$0 \to \ker f \to G \to G' \to 0$$
since we must have $\ker f \cong H$ ($g$ is injective so it's an isomorphism onto its image).
Kernels are normal subgroups, so you can indeed quotient by $H$.
You can make a similar argument for the image. Since it is surjective we can write $G' = \operatorname{Im}(f)$ and
$$0 \to \ker f \to G \to \operatorname{Im}(f) \to 0$$
And it is now clear that the first isomorphism theorem applies.
A: I usually interpret short exact sequences to be quotients (or extensions) like that - although they exist in much more general contexts. The one is supposed to denote the trivial group (ie. so the kernel of $g$ is trivial). In contexts other than group theory, $0$ is used instead.
As such I'll denote $1$ the trivial group. If $$1 \to H \stackrel{g}\to G \stackrel{f}\to G' \to 1$$ is exact, then it is exact at any object in the sequence. Note that the sequence being exact at $H$ is equivalent to $g$ being injective and exactness $G'$ is equivalent to $f$ being surjective. This is because there are trivial homomorphisms $1 \to H$ and $G' \to 1$ implicit in the diagram.
Since $g$ is injective, we know $H \cong \text{im}\;g$. That's why it's common to take $H$ to be a subgroup of $G$ without loss of generality. I'll avoid any of that so that there will be a clear distinction between equality and isomorphism.
So next, exactness at $G$ means that $\text{im}\;g = \ker f$ (this is strict equality). Then, using this equality, we note that $f: G \to G'$ is surjective so we can apply the first isomorphism theorem to find $G' \cong G/\ker f = G/\text{im}\;g$
