Answer to Question #105105 in Abstract Algebra for Sourav Mondal

Question

Check whether or not R is a normal subgroup of the group H = (R + Ri + Rj+ Rk,+ )where i²=j²=k² =-1,ij=-ji,jk=-kj,ki=-ik=and
( a+ bi + cj+ dk) + (a ′ + b′i + ′jc + d′ )k = (a + a′) + ( b+ b′ i) + (c + c′ j) + ( d+ d′ k) for
d,c,b,a,d',c',b',a ′∈ R.

Answer 1

Clearly,$0\in\R$ .let $a,b\in\R$ then $a+(-b)=a-b\in\R$ .

Hence ,by one step subgroup test $\R$ is a subgroup of $H$

Let $r$ be any element of $H.$

Then, $r+\R=\R +r \space, \forall \space r \in H$

Since,$a+bi+cj+dk+x=a'+b'i+c'j+d'k+x$ $\forall \space a,b,c,d$ $a',b',c',d',x\in\R$ .

Hence ,$\R$ is a normal subgroup of $H$ by the definition of normal subgroup