## Stream: maths

### Topic: normal subgroup : group :: ?? : semimodule

#### Kenny Lau (Jul 06 2020 at 08:42):

Normal subgroups are the kernels of group homomorphisms. More precisely, a subset of a group is a normal subgroup iff it is the kernel of some group homomorphism.

Then how do we characterise kernels of semimodule homomorphisms? Obviously the kernel of a semimodule homomorphism is a submodule; but given any submodule can we produce a homomorphism with the specified kernel?

#### Kenny Lau (Jul 06 2020 at 08:44):

For example, consider $(a,b) \mapsto a-b: \mathbb{N}^2 \to \mathbb{Z}$, a homomorphism between $\mathbb{N}$-semimodules. Its kernel is $\{(n,n) \mid n \in \mathbb{N}\}$.

#### Kenny Lau (Jul 06 2020 at 08:45):

Given this kernel, how do we recover $\mathbb{Z}$?

#### Kenny Lau (Jul 06 2020 at 08:47):

Maybe given kernel $S$ we should define $a \sim b := \exists s \in S, b=a+s$

#### Kenny Lau (Jul 06 2020 at 08:47):

good luck proving that this relation is symmetric

#### Kenny Lau (Jul 06 2020 at 08:48):

what if we define $a \sim b := \exists s \in S, b=a+s \lor a=b+s$ (good luck proving transitivity)

#### Chris Hughes (Jul 06 2020 at 08:48):

The kernel is a relation. Two homomorphisms with the same kernel do not have isomorphic images necessarily. e.g. The multiplicative map from N->N sending even numbers to zero and odd numbers to themselves has kernel {1} but is not the identity.

#### Kenny Lau (Jul 06 2020 at 08:50):

aha, we're in multiplicative land

#### Kenny Lau (Jul 06 2020 at 08:51):

ok, then what are n and s conditions on the relation to ensure that it is the kernel of some map?

#### Chris Hughes (Jul 06 2020 at 08:52):

It has to be a subsemimodule of the product.

#### Chris Hughes (Jul 06 2020 at 08:53):

i.e. it must "preserve" all the operations so I can define them on the quotient.

#### Chris Hughes (Jul 06 2020 at 08:54):

@Amelia Livingston did all this for monoids. Not for semimodules yet.

#### Chris Hughes (Jul 06 2020 at 09:01):

multiset.card is a better example of a map with trivial kernel that's not injective.

#### Kevin Buzzard (Jul 06 2020 at 09:35):

The additive map from nat to fin (n+1) sending all big numbers to n is another example (where the kernel doesn't determine the map). Amelia taught me how to think about this, although doubtless it was well known beforehand. A surjection of sets/types $X\to Y$, up to isomorphism, is just an equivalence relation on $X$. And now when you start adding structure, a surjection of groups/rings/whatever is just an equivalence relation which plays well wrt the structure. Analysing the lattice of normal subgroups of a group is just the same as analysing the lattice of isomorphism classes of quotients of the group. The quotient story is the one which generalises

#### Chris Hughes (Jul 06 2020 at 09:57):

This came up in my Group Theory course. There was this yucky definition of "block"s of a group action. Let $G$ be a group acting on a set $\Omega$. A non-empty subset $B$ of $\Omega$ is said to be a block if for each $g \in G$, either $gB=B$ or $gB \cap B = \emptyset$. Everything became much easier when I realised a block is just an equivalence class for an equivalence relation with the property $a \sim b \implies \forall g, ga \sim gb$, or equivalently the preimage of a point under a morphism of $G$-sets.

#### Yury G. Kudryashov (Jul 06 2020 at 18:35):

We have con and add_con. I hope eventually they will be used to define quotient groups etc

Last updated: May 18 2021 at 07:19 UTC