Question

Q. prove that
(i) H={a+ib ЄC, a2+b2=1} is a subgroup of C, Where C is complex number.
(ii) H be a set of real number a+b√2 where a,b ϵ Q. Show that H be a subgroup of non-zero real no. under *.

Answer 1

Answer on Question #60045 – Math – Abstract Algebra

Question

Prove that

(i) $H = \{a + ib \in C, a^2 + b^2 = 1\}$ is a subgroup of $C$, where $C$ is complex number.

(ii) $H$ be a set of real numbers $a + b\sqrt{2}$ where $a, b \in Q$. Show that $H$ be a subgroup of non-zero real no. under $*$.

Proof

(i) Let's prove that $H$ is a subgroup under multiplication.

If $r_1 = a_1 + ib_1$, $r_2 = a_2 + ib_2 \in H$, then

$$\begin{array}{l} r_1 * r_2 = (a_1 + ib_1) * (a_2 + ib_2) = a_1 * a_2 + ib_1 * a_2 + ia_1 * b_2 - b_1 * b_2 = \\ = a_1 * a_2 - b_1 * b_2 + i(b_1 * a_2 + a_1 * b_2), (a_1 * a_2 - b_1 * b_2)^2 + (b_1 * a_2 + a_1 * b_2)^2 = \\ = (a_1^2 + b_1^n z)(a_2^2 + b_2^2) = 1 \text{ and} \end{array}$$ $$r_1 * r_2 \in H.$$

Since $(a_1 - i b_1)(a_1 + ib_1) = a_1^2 + b_1^2 = 1$, then $r_1^(-1) = a_1 - ib_1$ and $r_1 \in H$. From these relations we obtain that $H$ is a subgroup.

(ii) If $r_1 = a_1 + b_1 \sqrt{2}$, $r_2 = a_2 + b_2 \sqrt{2}$, then

$$r_1 * r_2 = a_1 * a_2 + 2b_1 * b_2 + (a_1 * b_2 + b_1 * a_2) \sqrt{2} \in H.$$

Since $r_1^(-1) = -a_1 / (2b_1^2 - a_1^2) + b_1 / (2b_1^2 - a_1^2) \sqrt{2} \in H$

$$(r_1^(-1) * r_1 = 1 / (2b_1^2 - a_1^2) (-a_1 + b_1 \sqrt{2})(a_1 + b_1 \sqrt{2}) = 1 / (2b_1^2 - a_1^2)$$ $$(2b_1^2 - a_1^2) = 1, \text{ then } H \text{ is a subgroup under } *.$$

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