Determining the symmetry group of an infinite horizontal line. I believe I have a satisfactory answer to the following question:

Imagine we have a infinite horizontal line running through the origin, what is the associated symmetry group?

Now thinking intuitively about the symmetries of this line we can see it is invariant under:
Identity (obviously).
Rotation by $\pi$.
Reflection along the $x$ axis.
Reflection along any line $x=a\;,\; a\in \mathbb{R}$.
Translation by any $b \in \mathbb{R}$ along the $x$ axis.
Now the first four form the group $V_4$ (though not sure whether I can class the reflection along $x=a$ as one element?) and then adding in the translations we end up with a semi-direct product group (as matrix multiplication and addition of scalars don't commute) giving $$V_4\ltimes \mathbb{R}$$ I am also not sure if the above expression tells someone that the translation can only be done in the $x$ direction. Any thoughts?
 A: The group of affine transformations of a real line into itself, called the general affine group of the line, is
$$
\text{Aff}(1,\Bbb{R}) = \Bbb{R} \rtimes \text{GL}(1,\Bbb{R}) \simeq \Bbb{R} \rtimes (\Bbb{R}\setminus \{0\})
$$
where $\Bbb{R}$ represents translations and $\text{GL}(1,\Bbb{R})$ represents the affine transformations that leave the origin fixed, namely the group generated by the reflection along the $y$ axis (which is the same as the rotation by $\pi$) and the dilations. Indeed, a dilation on a real line can be represented as multiplication by a positive real number, while the rotation by $\pi$ can be represented by multiplication by $-1$.
If, instead, you are looking for the group of isometries of the line into itself, i.e. the affine transformations that preserve lengths, all you need to do is rule out the dilations, obtaining
$$
\Bbb{R} \rtimes \{1,-1\}
$$
where $\{1,-1\}$ is considered as the cyclic group of order $2$ in multiplicative notation.

Concerning your last question: note that the direction in which you are taking translations is implicit in the isomorphism. Indeed, you are identifying the translation by $a \in \Bbb{R}$ along the $x$ axis with the number $a$ itself. Furthermore, this isn't really a source of confusion, because any other translation of the plane in which you are embedding your line (you could even consider the line by itself) doesn't send the line to itself, thus it can't a symmetry of the line.

Finally, let's try to provide an explicit isomorphism for the case of the group $G$ of isometries of your line into itself. As you noticed, the only elements of $G$ are the translations $\tau_a$ by $a \in \Bbb{R}$ (note that the identity $\text{id}$ is the translation by $0$) and the reflections with respect to some vertical line $y = b$ with $b \in \Bbb{R}$.
Now, note that the rotation $\rho$ by $\pi$ coincides with the reflection with respect to the $y$ axis, i.e. the line $y = 0$. Furthermore, every other reflection by a vertical line can be obtained as the composition $\tau_a \circ \rho \circ \tau_{-a}$. Thus we can generate $G$ with the translations and with $\rho$.
Clearly the set of translations $T$ is a subgroup, and it is normal because
$$
\rho \circ \tau_a \circ \rho = \tau_{-a}
$$
for every $a \in \Bbb{R}$.
On the other hand, the subgroup $R = \{\text{id},\rho\}$ isn't normal because
$$
\tau_b \circ \rho \circ \tau_a \notin R
$$
for every $a,b \in \Bbb{R} \setminus \{0\}$. Finally observe that $T \cap R = \{\text{id}\}$ because $\rho$ isn't a translation, thus
$$
G \simeq T \rtimes R \simeq \Bbb{R} \rtimes \{1,-1\}.
$$
A: If you think to the line as a vector space $\mathbb{R}$ (not a subspace of some $n$-dimensional space),  than the symmetry group is the group of isometries, i.s. the group of linear transformations that conserve distances. We can easily see that all linear transformations are represented by $y=ax$ and the isometries are such that $|a|$=1. So we have two isometries: $y=x$ ( the identity) and $y=-x$ (symmetry with respect to the origin).
Not that the only rotation defined on $\mathbb{R}$ is the identity.
If you consider the line as an affine space than you have to add translations, i.e. transformations $y=x+a$.
