Show that Cayley's theorem, when applied to $\mathbb{R}$, produces the subgroup of $S_{\mathbb{R}}$ which contains all translations of the real line this is on the Groups and Symmetry textbook by Armstrong. I am completely lost with this question, and i'm unsure of the way to even begin with this question. Would someone help to give me a nudge in the right direction?
Or with any question that requires me to prove isomorphism using Cayley's theorem.
 A: Presumably, he means the group $(\Bbb R,+)$.
Recall that for a group $G$, one explicit construction of Cayley's Theorem is what is known as the left regular action: $\phi: G \to S_G$, given by:
$g \mapsto L_g$, where $L_g: G \to G$ is the bijection that acts on $a \in G$ 
 as $L_g(a) = ga$ (the map $L_g$ is also called "left-translation by $g$").
This is a homomorphism, since $\phi(gh) = L_{gh} = L_g \circ L_h = \phi(g) \circ \phi(h)$, as a consequence of the associativity of multiplication in $G$:
$L_{gh}(a) = (gh)a = g(ha) = L_g(ha) = L_g(L_h(a)) = (L_g \circ L_h)(a)$, for any $a \in G$.
Moreover, this is a faithful action (that is, $\phi$ is injective), since if $\phi(g) = \text{id}_G$, that is:
$ga = a$, for all $a \in G$, then $g = e_G$. Hence $G \cong \phi(G) \subseteq S_G$, by the First Isomorphism Theorem.
With the operation of addition of real numbers, it is more common to refer to $L_x$ for $x \in \Bbb R$, as $T_x$, where:
$T_x(y) = x + y$, for $y \in \Bbb R$. This is known as "translation by $x$", since it takes $y$ on the real line, and "moves it over by $x$". I believe you can fill in the missing gaps, now.
