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:

enter image description here

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?

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    $\begingroup$ He's not saying that each map $T_a$ is a homomorphism, but instead the map $T:G\to\mathfrak{S}_G$, where $\mathfrak{S}_G$ denotes the set of permutations of $G$ (which is a group by composition), given by $T(a)=T_a$, is a homomorphism. See that he wrote "so $T_{ab}=T_aT_b$", which means that $T$ is a homomorphism. $\endgroup$ – Luiz Cordeiro May 28 '15 at 4:30
  • $\begingroup$ @LuizCordeiro I find the notation $T_a T_b$ quite confusing. I think its the same as $T_{a * b} = T_a *' T_b$ where $*'$ denote the group operation on the set of $T_a$'s. What is confusing to me is, is the set $ \mathcal{G}_G$ a set of functions or are they elements of G? i.e. is it the same as G? I thought thats what a permutation was, the same elements but permuted around. $\endgroup$ – Pinocchio May 28 '15 at 5:30
  • $\begingroup$ @LuizCordeiro I guess what is not clear to me is what "the group of permutations of the set $G$" entails (in a precise rigorous way). $\endgroup$ – Pinocchio May 28 '15 at 5:39
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    $\begingroup$ If $X$ is any set, a permutation of $X$ is a bijection $f$ from $X$ to itself. We form the set $\mathfrak{S}_X$ of all permutations of $X$. This has a natural operation, namely composition: Given two permutations $f,g:X\to X$, we denote $f\circ g$, or simply $fg$, the map $fg:X\to X$ given by $fg(x)=f(g(x))$ for all $x\in X$. This gives a group structure on $\mathfrak{S}_X$ (indeed, the identity on $X$ is the unit of this group, and every bijection has an inverse). Does that solve your problem? $\endgroup$ – Luiz Cordeiro May 28 '15 at 6:46
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    $\begingroup$ When dealing with groups of permutations, we (essentially) always assume the operation is composition. Note, however, that when $G$ is a group and $X$ is any set, we can form the set $G^X$, consisting of all functions $f:G\to X$ from $G$ to $X$, and this has a group structure given in the following way: For $f,g:X\to G$, we define a new function $f*g:X\to G$ by $(f*g)(x)=f(x)*g(x)$ for all $x\in X$. This defines a group structure on $G^X$. However, this group structure is completely different from the group structure on the permutations of $G$, as described above. $\endgroup$ – Luiz Cordeiro May 28 '15 at 6:56

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 $n$.

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    $\begingroup$ why was this down voted? $\endgroup$ – Pinocchio May 28 '15 at 4:37
  • $\begingroup$ What is $S(G)$? A permutation is just an element of G, so why isn't this a map from G to G, :/ $\endgroup$ – Pinocchio May 28 '15 at 5:27
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    $\begingroup$ @Pinocchio A permutation is not an element of $G$, it is a bijective map from $G$ to itself. $\endgroup$ – Tobias Kildetoft May 28 '15 at 6:28
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    $\begingroup$ To make sure I actually understand this, is $S(G) = \{ \pi: G \rightarrow G \}$ the set of all functions $\pi$ that are permutations on $G$? Therefore $|S(G)| = |G| ! = n! $, right? $\endgroup$ – Pinocchio May 28 '15 at 17:37
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    $\begingroup$ Yes, that's right $\endgroup$ – Zach Effman May 28 '15 at 17:52

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:

$$T(a * b) = T(a) \circ T(b) $$


$$T(a * b) (x) = T(a) \circ T(b) (x) = T_{ab}(x) = T_{a} \circ T_{b}(x) $$

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:

  1. Closed: Obviously, if we compose two permutations we get another permutation in $T'$ (that preserves the bijective property).
  2. Associative: the order or parenthesis clearly doesn't matter when we compute a permutation.
  3. Identity: the identity permutations leaves any permutation unchanged hence $T_a \circ E = E \circ T_a = T_a $ holds true.
  4. 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! :)

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    $\begingroup$ One minor issue: I would say that it is more natural to let the codomain of your mapping be the set $S(G)$ of all bijections on $G$. Not all of these will be of the form $T_a$, as for a finite group there are $n!$ permutations but only $n$ elements to make $T_a$'s out of. Your set $T'$ is $Im(T)$, the image of the homomorphism, which sits inside of the whole set of permutations and is strictly a subset of them. $\endgroup$ – Zach Effman May 28 '15 at 17:29
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    $\begingroup$ Everything else is correct though, and it is perfectly correct to make the choice for $T'$ as you have. However, if you let $T'$ sit inside $S(G)$, you obtain a method to embed an isomorphic copy of $G$ inside the group of permutations on the elements of $G$, which is an interesting result known as Cayley's theorem. $\endgroup$ – Zach Effman May 28 '15 at 17:36
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    $\begingroup$ @ZachEffman I see. You are right $T'$ doesn't cover all the permutations. But, that is fine no (in the context of this question)? because $T_a$ (the function in question) can't possibly cover all permutations since its indexed by $G$. Or what would be the change to my answer to make it 100% correct. I care that I actually understand this and not tricking myself into thinking I do when I don't. Btw, thanks for your help! :) $\endgroup$ – Pinocchio May 28 '15 at 17:39
  • $\begingroup$ @ZachEffman, I did do a change to my answer, I (corrected) and explicitly said that $T'$ isn't the set of all permutations but a subset, that should make it correct now I think. $\endgroup$ – Pinocchio May 28 '15 at 17:42
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    $\begingroup$ I think you were correct to begin with, I just wanted to be sure you understood that $T'$ couldn't be all the permutations, and it's clear that you did $\endgroup$ – Zach Effman May 28 '15 at 17:55

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