Let $H$ be the Division ring of Quaternions. Let $\alpha =a + bi +cj + dk \in H$, we will call $\bar{\alpha} = a - bi - cj - dk$ the conjugate of $\alpha$.
Denote $Tr(\alpha) := \alpha + \bar{\alpha} = 2a, \text{ and } N(\alpha) := \alpha.\bar{\alpha} = a^2 +b^2 +c^2 +d^2$
Let $G_0 := \{\alpha \in H^* | N(\alpha)=1\}.$ Prove the following:
(1) $Tr(\alpha \beta) = Tr(\beta \alpha)$ và $N(\alpha \beta) = N(\alpha)N(\beta)$, for all $\alpha, \beta \in H$.
(2) $G_0$ is a subgroup of $H^*$ and $G_0$ contains the commutator subgroup $H' = [H^*, H^*]$ of $H^*$, where H^* = H{0}.
(3) Prove that $\alpha$ and$\beta$ are conjugates ( if and only if $Tr(\alpha)=Tr(\beta)$ and $N(\alpha)=N(\beta)$.
I solved the question (1) and the first half of (2). I need your help at question (2): the part where you have to show that $H'\leq G_0$. Thank you in advance.
Here is my work so far:
(1) Let $x = a + bi +cj + dk , y =e + fi +gj + hk,$ where $a,b,c,d,e,f,g,h \in R$, be elements of $H.$ We have $ x.y = ae - bf - cg - dh + (af + be + ch - dg)i + (ag - bh + ce + df)j + (ah + bg - cf + de)k$.
$y.x = ea - fb - gc - hd + (eb + fa + hd - gc)i + (ec - fd + ga + hb)j + (ed + fc - gb + ha)k $.
Hence $Tr(xy)=2ae - 2bf - 2cg - 2dh=Tr(yx)$, and $N(xy)=(ae - bf - cg - dh)^2 +(af + be + ch - dg)^2 +(ag - bh + ce + df)^2 +(ah + bg - cf + de)^2 = (a^2 +b^2 +c^2 +d^2)(e^2 +f^2 +g^2 +h^2) = N(x)N(y).$
(2) To prove that G is a subgroup of H*, we need to show that G satisfies three conditions: closure under the group operation, existence of an identity element, and existence of inverses for each element.
- Closure under the group operation:
Let x and y be two elements in G, which means that $N(x)= 1$ and $N(y) = 1$. We want to show that $xy$ is also in G. By the above, we have $N(xy) = N(x).N(y) = 1.1 = 1$. Therefore, $xy \in G,$ and G is closed under the group operation.
- Existence of an identity element:
Let $e = 1 + 0i + 0j + 0k$. Then $N(e) = 1^2 + 0^2 + 0^2 + 0^2 = 1$. Therefore, the identity element e is in G.
- Existence of inverses:
For every element $x$ in G, we need to show that there exists an element $y$ in G such that $xy = yx = e$, where e is the identity element. Let $x$ be an arbitrary element in G, and let $y= a - bi - cj - dk.$ Since $N(x)=1$ we also have $N(y)=1$.
Now we have: $xy = (a + bi + cj + dk)(a - bi - cj - dk) = a^2 - abi - acj - adk + abi + b^2i^2 + bcji + bdki + acj + bcj^2 + cdji + cdkj + adk + bdkj + cdkj + d^2k^2 = a^2 + b^2i^2 + c^2j^2 + d^2k^2 = a^2 + b^2 + c^2 + d^2 =N(x)=1.$
Therefore, xy = 1, and y is the inverse of x in G.
Since G satisfies all three conditions, namely closure under the group operation, existence of an identity element, and existence of inverses, we can conclude that G is a subgroup of H*.$