Is the left translation $T_a(x) =ax $ a homomorphism? I apologize if this is a super basic question but I was reading Lang's undergraduate algebra book and it says that the following function is a homomorphism:
$$T_a(x) = ax$$
The way I would check if something is a homomorphism is by checking if the algebraic structure of the original group is preserved. So say we have a group $(G,*)$ and we want to show $T_a$ is a homomorphism. The way I would do it is by checking if the following equations holds:
$$f(x*y) = f(x) *' f(y)$$
in this case:
$$T_a(x * y) = T_a(x) * T_b(x) $$
So lets check:
LHS
$$T_a(x * y) = a * x * y$$
RHS
$$T_a(x)*T_a(y) = a * x * a * y$$
which doesn't seem to be the equation that we wanted to hold true. So for me its clear its not a homomorphism...
However, Lang argues it is with the following which I do not understand at all:

We contended that it is a homomorphism. Indeed, for $a, b \in G$ we have
  T_{ab}(x) = abx = T_a( T_b(x) )
  so that $T_{ab} = T_aT_b$

I will provide a screenshot to of the whole thing to make sure I didn't miss out any important detail:


Also, after seeing the comment and the current answer, I more or less see what my confusion is but I am still confused.
In particular, what does the sentence:

"... the group of permutations of the set $G$."

means rigorously?
In particular, what does:
$$T_{ab} = T_{a}T_{b} $$
mean rigorously?
Does it mean:
$$T_{a * b}(x) = T_{a}(x) *' T_{b}(x) $$
with $(G, *)$ and $(T, *')$ respectively. (note $T = \{ T_a = T_a( \cdot ): a \in G \}$ is the group of functions that map according to $T_a(x) = ax$.)
or does it mean:
$$T_{a * b}(x) = T_{a} *' T_{b}(x) =  T_{a} \circ T_{b}(x) = T_{a}(T_{b}(x))$$
whichever it means, can someone explain me why it means that?
 A: The author is not arguing that translation is a homomorphism. Rather, he is showing that left-translation is a bijection, and hence a permutation of the group. He then shows that the map $G \to S(G)$ given by $a \to T_a$ is a homomorphism. Note that the homomorphism is from $G$ to $S(G)$, the group of permutations of $G$; the permutations in the image of the homomorphism are not homomorphisms of $G$ to $G$.
Added: The group of permutations of a set $S$ is the set of bijections from $S$ to $S$ with the group operation being function composition and the identity element being the identity function. It is easy to show that these functions form group, because composing bijections gives a bijection, and bijections have bijective inverses. As an example, the standard symmetric groups $S_n$ correspond respectively to the group of permutations of finite sets of size $n$.
A: I think I understand now. The main confusion was that that it said that the map:
$$ a \rightarrow T_a$$
is a map from $G$ into the group of permutations of the set $G$. The set of permutations is itself a function, because any permutation on the group $G$ is specified by the mapping (i.e. the permutation). In that case I will use the notation $(T', *') = (T', \circ)$ to denote a subset of the group of permutations $S(G)$ and the operation $ *' = \circ$ will be composition (where $T' = \{ T_a: G \rightarrow G \mid a \in G, T_a(x) = ax\}$ is the set of permutations). I will also use $T_a = T(a)$ to select a specific permutation function. In that case we can think of the homomorphism the text is talking about as follow is:
$$T: (G, * ) \rightarrow (T, \circ)$$
where we want to check the usual property $T(x * y) = T(x) *' T(y)$ in this context we want to confirm:
$$T(a * b) = T(a) \circ T(b) $$
or
$$T(a * b) (x)  = T(a) \circ T(b) (x) = T_{ab}(x) = T_{a} \circ T_{b}(x) $$
Lets check this:
$$T_{a * b}(x) = a * b * x = a * (b * x) = a * T_b(x) = T_a( T_b(x) ) = T_a \circ T_b (x)$$
which is exactly what we needed $T(a * b) = T(a) \circ T(b) $ or $T_{a * b} = T_{a} \circ T_{b}$. Therefore the map from the group $G$ to the group of permutations $T'$ is a homomorphism as required!

As a side comment its nice to confirm that $(T', *') = (T', \circ)$ is indeed a group under composition. I will discuss it intuitively and briefly:


*

*Closed: Obviously, if we compose two permutations we get another permutation in $T'$ (that preserves the bijective property).

*Associative: the order or parenthesis clearly doesn't matter when we compute a permutation.

*Identity: the identity permutations leaves any permutation unchanged hence $T_a \circ E = E \circ T_a = T_a $ holds true.

*Inverses: For any permutation, its easy to convince yourself that since its a bijective function, it clearly has a bijective inverse.


So the set of permutation functions under composition is a group! :)
