Question

b) Let a R3 be a vector of length 1. Define H={x E R3 : a·x=0). Here a x denotes the dot product of the vectors a and x. (i) Show that H is a subgroup of R (ii) For λ E R, show that : a·x= is a coset of H in R3. (ii) Is H cyclic? Prove or disprove.

Answer 1

b. (i) Recall to show H is a subgroup of $\mathbb R^3$, it is enough to show for any $x,y\in H$, $x-y\in H.$

Now note that $x,y\in H$ implies $a.x=a.y=0$.

Now $a.(x-y)=a.x-a.y=0$, hence subgroup.

(ii) Recall coset of H will looks like , for some $g\in \mathbb R^3$ . So to prove $H_{\lambda}$ is a coset of H we need to show $H_{\lambda}$ is of the form for some $g\in \mathbb R^3$ .

Take $g=\lambda a\in \mathbb R^3$ . Then note that $H_{\lambda}=g+H$ , as clearly $g+H\subset H_{\lambda}$ , as for any $x\in H$, $a.(g+x)=a.\lambda a+a.x= \lambda |a|^2 +0=\lambda$ , and for any $x\in H_{\lambda}$ , write $x=g+(x-g)$, and note that $a.(x-g)=a.x-a.g=\lambda-\lambda|a|^2=\lambda-\lambda=0$ , hence $x-g\in H$, hence $x\in g+H$, hence the equality.

(iii) Note that H can not be cyclic as take $a=(1,0,0)$, then note that and $(0,0,1)\in H$. Then note that if H is cyclic then there exists y n H such that $ny=(0,1,0)$ and
my-(0,0, 1
, for some integer n, m. This gives us the as a vector space $(0,0,1), (0,1,0)\in L\{y\}$ , where is the linear span of y, as a $\mathbb R^3$ vector space. Thus which implies are linearly dependent which is a contradiction. Hence H can not be cyclic.

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