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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.

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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.

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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.

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  • $\begingroup$ This is extremely helpful, thank you. Would you mind explaining what you mean by "unique up to unique isomorphism"? What is the difference between "unique up to isomorphism" and "unique up to unique isomorphism"? $\endgroup$ – user861776 Apr 11 at 20:13
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    $\begingroup$ @user861776 The former means that there is some non-zero number of isomorphisms between the various options. "Unique up to unique isomorphism" means that there is exactly one isomorphism. Equivalently, it's unique and has no non-trivial automorphism. For example, there is one group of order 2 up to unique isomorphism (any isomorphism has to fix the identity element, and hence the non-identity element), but the one group of order 3 is not unique up to unique isomorphism, because there's a non-trivial choice in defining the isomorphism (the inverse map is a non-trivial automorphism). $\endgroup$ – user3482749 Apr 11 at 21:56
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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 $G' \cong G/\ker f = G/\text{im}\;g$

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