Math 2.0

example

A translation of the integers is not a homomorphism

Example

Let G=(Z,+)G = (\mathbb{Z}, +) be the integers with addition. Define t:G→Gt : G \to G by t(u)=u+1t(u) = u + 1. A homomorphism sends the identity to the identity. But t(0)=1t(0) = 1, and 11 is not the identity 00. We can also compute directly: t(0+0)=1t(0 + 0) = 1, but t(0)+t(0)=2t(0) + t(0) = 2. Thus tt is not a homomorphism.

Remarks

tt is a bijection, so a bijection between groups need not be an isomorphism.

What this node needs for W1
  • an inventory row: the row of the inventory that the node accounts for (0 of 1)

The writing levels say how complete the text of a node is. They do not measure its truth.

What this needs

To read this node, read these 4 nodes in this order. The foundations come first, and the list ends with this node.

Group, concept, no evidence levelGroupconceptIntegers, object, no evidence levelIntegersobjectGroup homomorphism, definition, no evidence levelGroup homomorphismdefinitionA translation of the integers is not a homomorphism, example, no evidence levelA translation of the …example
Each node uses the nodes above it. The colour of the bar on a box shows the evidence level, and the box also gives it as text. The list below has the same nodes.
  • no evidence level
  1. Group concept
  2. Integers object
  3. Group homomorphism definition
  4. A translation of the integers is not a homomorphism example

Sources

  1. Web https://en.wikipedia.org/wiki/Group_homomorphism, Properties

What links here

No node links to this node yet.