Clearly,0∈R .let a,b∈R then a+(−b)=a−b∈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 ,∀ r∈H
Since,a+bi+cj+dk+x=a′+b′i+c′j+d′k+x ∀ a,b,c,d a′,b′,c′,d′,x∈R .
Hence ,R is a normal subgroup of H by the definition of normal subgroup
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