Let $\mathbb{C}^{\times}$ be the group of nonzero complex numbers under multiplication. Then $\mathbb{C}^{\times}$ is the direct product of the circle group $T$ of unit complex numbers and the group $\mathbb{R}^{+}$ of positive real numbers under multiplication.
[Math] Prove the direct product of nonzero complex numbers under multiplication.
abstract-algebracomplex numbers
Related Solutions
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* \cong \newcommand{\Z}{\mathbb{Z}}$ $\Z/n\Z \oplus \bigoplus_{i=1}^{\infty} \Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
We first show that $G = \mathbb{C}^* = \mathbb{C} − \{0\}$ under complex multiplication forms a group.
- Closure:
Let $z = a + bi$ and $w = c + di$ be both in $\mathbb{C}^∗$. Then we have $zw = (a + bi)(c + di) = ac − bd + (ad + bc)i$. To show that this element is in $G$, we compute $$(ac − bd)^2 + (ad + bc)^2 = a^2c^2 − 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2c^2 \\= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 \\= a^2(c^2 + d^2) + b^2(c^2 + d^2) \\= (a^2 + b^2)(c^2 + d^2).$$ Now $z, w ∈ G$ imply that a $a^2+b^2 > 0$ and $c^2+d^2 > 0$, whence $(ac−bd)^2+(ad+bc)^2 > 0$ and $zw ∈ G$.
- Associativity:
To show associativity, let $z = a + bi, w = c + di$ and $u = e + fi$. Then
$$(zw)u = [(a + bi)(c + di)](e + f i) = [(ac − bd) + (ad + bc)i](e + f i) \\= [(ac − bd)e − (ad + bc)f] + [(ac + bd)f + (ad + bc)e]i \\= (ace − bde − adf − bcf) + (acf − bdf + ade + bce)i \\= [(a(ce − df) − b(cf + de)] + [a(cf + de) + b(ce − df)]i \\= (a + bi)[(ce − df) + (cf + de)i] \\= (a + bi)[(c + di)(e + f i)] \\= z(wu).$$
- Identity Element and Inverse:
It is not difficult to check that $1 + 0i$ acts as a unit in multiplication and that
$$(a + bi)^{−1} = \frac{1}{a+bi} =\frac{a−bi}{(a+bi)(a−bi)} = \frac{a−bi}{a^2+b^2} = \frac{a}{a^2+b^2} − \frac{b}{a^2+b^2} i ∈ G$$ , since $$\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2 = \left(\frac{a^2+b^2}{(a^2+b^2)^2}\right) = \frac{1}{a^2+b^2} > 0$$
Since all the group axioms hold, $(G,\cdot)$, is a group under multiplication.
Ok, this is a group, now we let see that Circle Group is a Group.
Let $K$ be the set of all complex numbers of unit modulus: $K={z∈\mathbb{C}:|z|=1}$
- Closure:
Then the circle group $(K,\cdot)$ is an uncountably infinite abelian group under the operation of complex multiplication.
$$
z,w ∈ K
⟹ |z| = 1=|w|
⟹ |zw| = |z||w|
⟹ zw ∈ K $$
So $(S,\cdot)$ is closed.
Associativity, comes from complex multiplication is associative.
Identity : From Complex multiplication identity is one we have that the identity element of $K$ is $1+0i$.
Inverses
We have that $|z|=1⟹\frac{1}{|z|}=\left|\frac{1}{z}\right|=1$. But $z\cdot\frac{1}{z}=1+0i$. So the inverse of $z$ is $\frac{1}{z}$.
- Commutative: We have that Complex Multiplication is Commutative
So $K$ is a subgroup of $G$ under complex multiplication.
We can assert that from complex multiplication is commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group.
Best Answer
Define $f:{\mathbb C}^{\times}\to T\times\mathbb{R}^{+}$ by $f(z)=(\frac{z}{|z|},|z|)$ and prove that this is an isomorphism of groups.
Actually $\mathbb{C}^{\times}$ is the internal direct product of the two since ${\mathbb C}^{\times}=T\mathbb{R}^{+}$ (write $z=\frac{z}{|z|}\cdot|z|$) and $T\cap\mathbb{R}^{+}=\{1\}$.