1

Personally, I find left to right composition much more natural, but I have seem to come across an instance where using this notation gives different results.

Let $G$ be a group and define $\phi: G \to \operatorname{Aut} G$ as: $$ \phi : g \mapsto I_g $$ where: \begin{align} I_g : G &\to G \\ x &\mapsto g x g^{-1}\\ \end{align} Then $\phi$ is a homomorphism since: \begin{align} a,b \in G \implies \phi(ab) = I_{ab} = I_a I_b = \phi(a)\phi(b) \end{align} where $I_{ab} = I_a I_b$ follows from the fact that: $$ I_{ab} (x) = ab x b^{-1} a^{-1} = I_a (I_b (x)) $$ However, the preceding is not true if we use left to right notation since $(x) I_{ab} = (x) I_b I_a $ and not $(x) I_a I _b$. So if we use left to right notation, we instead have: $$ (ab) \phi = (b) \phi (a)\phi = I_b I_a $$


So, what gives? If notation can affect results, how can one justify the use of different notations?

Shaikh Ammar
  • 129
  • 1
    What gives is how you interpret $I_{ab}=I_aI_b$. The way I read it is that applying $I_{ab}$ is the same as applying $I_b$ first and then applying $I_a$ after. So in the new notation, $I_{ab}=I_b I_a$ is unsurprising since you're applying $I_b$ first and then applying $I_a$ after. So to make how one reads right to left composition consistent, one must have $I_{ab}=I_b I_a$. – Shrugs Nov 26 '23 at 21:33
  • 1
    This is possibly related to this old question. – HallaSurvivor Nov 26 '23 at 22:39

1 Answers1

3

The following statement is independent from any definition of the composition $\circ$:

If $G$ is a group, then $\mathrm{Aut}(G)$ is a group with multiplication defined by $(\alpha \cdot \beta)(g) := \alpha(\beta(g))$. Also, conjugation provides a homomorphism $G \to \mathrm{Aut}(G)$.

The following statement is dependent on a definition of the composition $\circ$:

If $G$ is a group, then $\mathrm{Aut}(G)$ is a group with multiplication defined by $\alpha \cdot \beta := \alpha \circ \beta$. Also, conjugation provides a homomorphism $G \to \mathrm{Aut}(G)$.

As you can see, this is not related to the conjugation map. It is really about the definition of the multiplication in the group $\mathrm{Aut}(G)$. When you are using the more common definition of $\circ$, the statements are the same. But when you are using the other "left to right" definition of composition, which is more commonly denoted as $\,;\,$ at least in computer science circles (also in order to avoid any confusion with $\circ$), then $a \cdot \beta := \alpha \,;\, \beta$ actually yields the opposite group $\mathrm{Aut}(G)^{\mathrm{op}}$ (for the usual definition of $\mathrm{Aut}(G)$), and the second statement will not be true anymore when $\circ$ is just replaced with $;$.

Given this, I am not convinced that any statements change their meaning when you change the notation. You just gotta be very careful to change the notation consistently. Here, you forgot to also change it in the definition of $\mathrm{Aut}(G)$. When you want to get a homomorphism $G \to \mathrm{Aut}(G)$, you need to define $\alpha \cdot \beta := \beta \,;\, \alpha$ (which is $\alpha \circ \beta$).

For a more drastic example (just to illustrate the point): Let's say I define natural number literals by $-1 := S(0)$, $-2 := S(S(0))$, $-3 := S(S(S(0)))$, etc., where $S$ is the successor function. So $\{\dotsc,-2,-1,0\}$ is the set of natural numbers. Now you might think: "Hey, the set of natural numbers doesn't have any smallest element anymore!". But you got fooled by the notation. $0$ is still the smallest element when you define $<$ as usual, because it will give you $0 < -1 < -2 < \cdots$.

Martin Brandenburg
  • 163,620
  • 1
    Of course, you can also get a homomorphism into the opposite automorphism group by letting $I_g : x \mapsto g^{-1}xg$. This will usually be the best place to put the flip, I think. – Eike Schulte Nov 27 '23 at 11:15