Introduction to Group Theory

Properties of Number Groups

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Recall the four axioms in the definition of a formal group; these will be tested on this page.

This page allows you to test the two most common binary operations (multiplication and addition) on the most common number sets (Integers, Rationals, Complex) for the axioms of the formal definition of a group. All three of these groups have closure, so that does not need to be shown.

Note that multiplication in $$\mathbb{C}$$ is defined as:

\require{amsmath} \begin{align*} (a+bi) \times (c+di) & = a(c+di) + bi(c+di) \\ & = ac + adi + bci + bdi^2 \\ & = ac + adi + bci - bd \\ & = (ac - bd) + (ad + bc)i \end{align*}

Close

$$\mathbb{Z}$$ = Integers, $$\mathbb{Q}$$ = Rationals, $$\mathbb{C}$$ = Complex

First, choose a number set:
Integers
Rationals
Complex

Next, choose whether to check it under addition or multiplication:
Addition
Multiplication

Now check the properties individually:


The complex numbers satisfy the group properties, but in a more "complex" way than do either the integers or the rationals. Here are the proofs for the inverse and associative properties for complex numbers:


Let there exist $$(a+bi) \in \mathbb{C}$$ and let there exist $$\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}$$. Now we multiply them together:

\require{amsmath} \begin{align*} (a+bi) \times \left[\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}\right] & = \left[\frac{a^2}{a^2+b^2} + \frac{b^2}{a^2+b^2}\right] + i\left(\frac{-ab}{a^2+b^2} + \frac{ba}{a^2+b^2}\right) \\ & = \frac{a^2+b^2}{a^2+b^2} + i\left(\frac{ab-ab}{a^2 + b^2}\right) \\ & = 1 + 0i \\ & = 1 \end{align*}

It is trivial to show that this property is commutative. Therefore, the inverse axiom is satisfied for C.

Q.E.D.

For $$(\mathbb{C},\times)$$:

\begin{align} \left[\left(a+bi\right)\times\left(c+di\right)\right]\times\left(e+fi\right) & = \left[\left(ac-bd\right)+i\left(ad+bc\right)\right]\left{e+fi\right) \\ & = \left[e\left(ac-bd\right)-f\left(ad+bc\right)\right]+i\left[f\left(ac-bd\right)+e\left(ad+bc\right)\right] \\ & = \left(ace-bde-adf-bcf\right)+i\left(acf-bdf+ade+bce\right) \\ \left(a+bi)\times\left[\left(c+di\right)\times\left(e+fi\right)\right] & = \left(a+bi)\times\left[\left(ce-df\right)+i\left(cf+de\right)\right] \\ & = \left[a\left(ce-df\right)-b\left(cf+de\right)\right]+i\left[a\left(cf+de\right)+b\left(ce-df\right)\right] \\ & = \left(ace-bde-adf-bcf\right)+i\left(acf-bdf+ade+bce\right) \end{align}

Q.E.D.

For $$(\mathbb{C},+)$$:

\begin{align} \left[\left(a+bi\right)+\left(c+di\right)\right]+\left(e+fi\right) & = \left(a+c+bi+di\right)+\left(e+fi\right) \\ & = a + c + e + bi + di + fi \\ & = \left(a + c + e\right) + i\left(b+d+f\right) \\ \left(a+bi\right)+\left[\left(c+di\right)+\left(e+fi\right)\right] & = \left(a+bi\right)+\left(c+e+di+fi\right) \\ & = a + bi + c + e+ di + fi \\ & = \left(a+c+e\right)+i\left(b+d+f\right) \end{align}

Q.E.D.