Isotropy subgroup of a non-zero vector in 3D space Recentely i got an assignment for a course of group theory applied to physics, in which i had to find the isotropy group of a vector $v=(1,0,0)^T$ under the euclidean group $E=\mathbb{R^3}\ltimes O(3)$. 
Our professor told us to neglect the translation group, meaning that $v'=A*v$ with $A \in O(3)$ instead of $v'=A*v+b$ with $b \in \mathbb{R^3}$. This of course makes the actual computation of the little group so much easier. So i was wondering if there is  a correct reason to neglect that term?
 A: Think about it, if one included translations and you still needed to stabilize  $ v = (1, 0, 0)^T$ then there would have to be a translation that did not move you in the z or y direction since those components in $v $ would be non-zero but it can't move you in the x either since that component will be different from $1 $. In the group of translation there is only one element that does this namely the identity translation. Therefore your subgroup $ E_H $ is really  $E_H = I \rtimes O(3)$ which is isomorphic to just $ O(3)$
A: A generic $(b,A)\in\mathbb R^3 \ltimes O(3)$ on a vector $v$ takes the form
$$v \mapsto A(v+b)$$
where $b\in \mathbb R^3$ and $A\in O(3)$.  If we're searching for the stabilizer $S$ of  $v=(1,0,0)^T$, then we seek transformations such that
$$\pmatrix{1\\0\\0} = A\pmatrix{b_1+1\\b_2\\b_3} \implies (A^{-1}-\mathbf 1)\pmatrix{1\\0\\0} = b$$
For a given $A\in O(3)$, let
$$b(A):= (A^{-1}-\mathbf 1) \pmatrix{1\\0\\0} $$
The group of elements $S=\big(b(A),A\big)$ constitutes a subgroup of $\mathbb R^3\ltimes O(3)$, as can be proven straightforwardly. It's also clear that $A \leftrightarrow \big(b(A),A\big)$ is a group isomorphism between $O(3)$ and $S$.

Clearly this procedure didn't actually depend on our choice of $v$.  If we actually apply one of the above transformations $\big(b(A),A\big)$ to $v=(1,0,0)^T$, we see that it can be understood as the application of a translation $v\mapsto \vec 0$, then an arbitrary orthogonal transformation to the zero vector, and then a final translation $\vec 0 \mapsto v$.
In other words, the stabilizer subgroup of $\mathbb R^3\ltimes O(3)$ corresponding to any vector $v$ is isomorphic to the stabilizer subgroup of the zero vector.

Our professor told us to neglect the translation group, meaning that $v'=A*v$ with $A \in O(3)$ instead of $v'=A*v+b$ with $b \in \mathbb{R^3}$.

That's not quite right. Neglecting the translation group means that you're analyzing the stabilizer corresponding to $\vec 0$, not $v=(1,0,0)^T$. If we ignore the translation subgroup and search only for those $A\in O(3)$ such that $v=Av$, we obtain a smaller group which is isomorphic to $O(2)$ (rotations and reflections in the plane orthogonal to $v$).
