In abstract algebra, Noether's isomorphism theorems are extremely general statements about algebraic objects and the relationships between them. I've studied the theorems in separate courses on groups, rings, and modules over commutative rings (including vector spaces), so I've spent plenty of time trying to grasp their essence. In this post I share some of the intuition that I've developed. If nothing else, this intuition helps to remember the statement of the theorems. I will present the theorems in the context of abelian groups, where the statements are especially straight-forward, in order to avoid clouding the intuition with details about normal subgroups and so on. I will number the theorems 1, 2, 3, 4.
Theorem 1. If $f$ is a homomorphism between abelian groups $A, B$ then $$\frac{A}{\text{Ker } f} \cong \text{Im } f$$
This is the core theorem from which theorems 2 and 3 easily follow. By taking the quotient of $A$ with the kernel of $f$, we are forming a group that looks like $A$ except where we've combined the elements that together map to the same element in the codomain. I think of this as forcing $f$ to become injective by taking all the sets of elements where it fails to be injective and combining those elements into a single element. Since our version of $f$ is now injective, and it maps onto its image (by definition of image), then obviously it's a bijection onto its image. Moreover, the structure-preserving nature of the homomorphism means that structure is preserved through the bijection – i.e. there is an isomorphism between the LHS and RHS.
To me, the statement becomes almost tautological under this interpretation. It seems to say "if we take a homomorphism and disregard all the points where it fails to be injective, then it becomes injective".
Theorem 2. If $A, B$ are subgroups of an abelian group $G$, then $$\frac{A+B}{A} \cong \frac{B}{A \cap B}$$
I think of this statement as a broad generalisation of a basic result from number theory: if $a, b$ are positive integers then $a/\gcd(a,b) = \mathrm{lcm}(a,b)/b$ (this rearranges into the more familiar form $ab/\gcd(a,b) = \mathrm{lcm}(a,b)$. If we draw the diamond-shaped partial order of the "divides" relation with the positive integers $a, b, \gcd(a,b), \mathrm{lcm}(a,b)$ and then we draw the partial order of the "subgroup" relation with groups $A, B, A \cap B, A + B$, we see that the number-theoretic statement on the "divides" partial order looks very similar to what the second isomorphism theorem says about the "subgroup" partial order.
Indeed, $A \cap B$ as the "greatest common subgroup" of $A, B$ and $A + B$ is the "smallest common supergroup" of $A, B$. We know that the quotient operation feels a lot like division, so it seems natural to me to relate the quotient operation with greatest common subgroups and smallest common supergroups to the division operation with greatest common divisors and least common multiples.
The number theory statement can in fact be derived from the second isomorphism theorem applied to the $(\mathbb Z, +)$, so we're not saying anything groundbreaking. The benefit of this view of things is that you probably already have a good intuition for the properties of GCDs and LCMs, so it's just a matter of translating that intuition to the world of abstract algebra.
This isn't a complete intuition. In particular it's not clear under this intuition why we get an isomorphism between the LHS and RHS rather than just a bijection. But it's a useful starting point, or at the very least a mnemonic device.
Theorem 3. If $A \leq B \leq G$ are abelian groups, then $$\frac{G/A}{B/A} \cong \frac{G}{B}$$
This is probably the easiest theorem to remember out of the three. Quotients feel like fractions, and indeed we can "multiply both sides" of the fraction to cancel out quotients of quotients. Unfortunately I don't yet have a good intuition for the preservation of structure.
It's worth keeping in mind that the proofs of the three theorems are all in the "follow your nose" style. For the first theorem we want to establish an isomorphism between the LHS and RHS; it turns out that the most natural possible mapping from the LHS to the RHS turns out to be this isomorphism. For the remaining theorems we again pick the most natural mappings between the LHS and RHS (it'll be easiest to take the mapping from the RHS to the LHS). Again this mapping will happen to be a homomorphism, and the second and third theorems fall out immediately from applying the first theorem to this homomorphism. This is quite remarkable and it means that even if you haven't developed a fully-formed intuition for the theorems, it's easy to convince yourself rigorously that they're correct.
Theorem 4. Let $P$ be a subgroup of an abelian group G. Then there exists a bijective correspondence$$\left\{ \text{subgroups of } \frac{G}{P} \right\} \longleftrightarrow \left\{ \text{subgroups of G that include P}\right\}$$
The right-to-left map looks like this: if $Q \leq P$ is a subgroup satisfying $P \subseteq Q$, then $P$ is also a subgroup of $Q$, so we can form the quotient $Q/P$. Since $Q \leq G$ it would seem natural that $Q/P \leq G/P$. Indeed this is the case, so $Q/P$ is in the LHS. In other words, we map from the right to the left simply by taking the quotient of the subgroup with $P$.
For the left-to-right map, notice that any subgroup of $G/P$ has elements of the form $g+P$ where $g$ is a coset representative. Denote the set of all such representatives by $H$. We can show that $H$ is a subgroup of $G$ (it inherits its structure from $G/P$). Moreover, the representatives of the identity element of the subgroup of $G/P$ are the elements of $P$ itself, so the elements of $P$ are contained within $H$. Thus $P \subseteq H \leq G$ so $H$ lands in the RHS. I visualise this map as "unpacking" a subgroup of the quotient into its representatives, and it makes intuitive sense to me that (1) these representatives form a subgroup of the quotient numerator, and (2) the identity of the quotient (i.e. the coset $0+P$) is in the subgroup and unpacks to the set $P$.
Some work has to be done to prove that these maps are inverses of each other, but if we think of the maps above as packing elements into a quotient and then unpacking the quotient back to elements, the fact that the maps are inverses of each other shouldn't come as a surprise.
No comments:
Post a Comment