# The exclusion of empty substructures

1. Dec 8, 2014

### 1MileCrash

So, subspaces of vector spaces, and subgroups of groups, are not allowed to be empty.

This is because "there exists an identity element". We could include the empty set in these substructures but have the definition otherwise unchanged.

I'm curious as to what the consequences of such would be. If the empty subset of a group G were considered a subgroup of G, what would be some consequences in our important theorems?

2. Dec 8, 2014

### mathman

All theorems which require a reference to the identity would have to include "except for the empty set", an unneeded complication.

3. Dec 11, 2014

### Fredrik

Staff Emeritus
A subgroup is supposed to be a subset that's also a group. Since the empty set is not a group, it would be pretty odd to insist on calling it a subgroup.

4. Dec 11, 2014

### 1MileCrash

Clearly, if the empty set were considered a subgroup, it would also be considered a group..

5. Dec 11, 2014

### Fredrik

Staff Emeritus
OK. But that means that we would have to change the definition of "group" from

A pair $(G,\star)$ is said to be a group if $\star$ is a binary operation on $G$ that satisfies the group axioms.​

to

A pair $(G,\star)$ is said to be a group if $G=\star=\varnothing$ or $\star$ is a binary operation on $G$ that satisfies the group axioms.​

This doesn't look like an improvement.

Some theorems would remain intact. For example, consider the theorem "For all $x,y,z\in G$, if $x\star z=y\star z$, then $x=y$." This statement is true when $G=\varnothing$, because $G$ doesn't contain three elements $x,y,z$ such that the implication is false.

6. Dec 11, 2014

### 1MileCrash

It's not really suggesting that it is an "improvement", I'm merely asking the question "what happens if we relax our axioms." We don't have to call this new object a group any more, it doesn't matter.

Immediately, Lagrange's Theorem will no longer work, for example, and G/{} would be a quotient group since {} is normal, and it seemingly would be the set {} again (the definition would lead {} to have no left cosets) but under the operation associated with quotient groups rather than that of G.

So a lot of things get weird or break right off the bat, but I'm wondering if anything more interesting would arise.

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