Question

Q. prove that H={a+ib C, a2+b2=1} is a subgroup of C. Where C is complex number. H be set of real number a+b where a,b Є ф. Show that H be a subgroup of non-zero real no. under multiplication.

Answer 1

Answer on Question #58603 – Math – Abstract Algebra

Question

Prove that $H = \{a + ib \in \mathbb{C} \mid a^2 + b^2 = 1\}$ is a subgroup of $\mathbb{C}$, where $\mathbb{C}$ is the set of complex numbers, $H$ is the set of numbers $a + bi$, where $a, b \in \varphi$. Show that $H$ is a subgroup of non-zero real numbers under multiplication.

Solution

It is given that $H = \{a + ib \in \mathbb{C} \mid a^2 + b^2 = 1\}$. Let $G = \{c + id \in \mathbb{C} \mid c^2 + d^2 = 1\}$. Then $H * G$ will be

$$(a + ib) * (c + id) = ac + iad + icb - bd = ac - bd + (ad + cb)i$$

We will prove that $(ac - bd)^2 + (ad + cb)^2 = 1$:

$$\begin{array}{l} (ac - bd)^2 + (ad + cb)^2 = a^2c^2 + b^2d^2 - 2abcd + a^2d^2 + 2abcd + b^2c^2 \\ = a^2(c^2 + d^2) + b^2(d^2 + c^2) = (a^2 + b^2)(c^2 + d^2); \end{array}$$

$(a^2 + b^2)(c^2 + d^2) = 1$, because $a^2 + b^2 = 1$ by the statement of the question and

$c^2 + d^2 = 1$ by the assumption.

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