Group theory - composition vs. multiplication

Question

In my professor's lecture notes, I've noticed that sometimes the $\circ$ symbol is used when operations are combined, but at other times multiplication is referred to , for example in the definition of a homomorphic relationship:

$$\phi:G\rightarrow H,\,\ \ \ g,h\in G, \ \ \ \phi(g), \phi(h)\in H$$ $$\phi(g\star h) = \phi(g)\cdot \phi(h)$$

Is there any difference between the two? To me it seems like the multiplication of $g$ and $h$ (e.g. in a symmetry group) simply represents two successive operations applied to an object, which is essentially the definition of function composition.

EDIT: I was perhaps a bit unclear here; my question was not directed at the difference between $\star$ and $\cdot$, it was the difference between composition and the group operation. I'll leave it as it is now, so that BobaFret and Foobaz John's answers make sense.

Here's a similar question, but that question deals with why we refer to the operation as multiplication, not whether it is analogous to composition (as far as I can tell).

Answer 1

The notation distinguishes between the binary operation for each group. $\star$ is the operation in $G$ and $\cdot$ is the operation in $H$. It stresses that $g \star h$ is in $G$ while $\phi(g) \cdot \phi (h)$ is in $H$.

Answer 2

The author seems to be stressing the fact that $\star$ and $\cdot$ are two different group operations. $\star$ is an operation on $G$ while $\cdot$ is an operation on $H$. Usually though we just write $$ \phi(gh)=\phi(g)\phi(h) $$ where the operation on the left is in $G$ and operation on the right is in $H$.

Answer 3

Function composition is defined as follows:

$$\circ\colon\operatorname{Hom}(Y,Z)\times \operatorname{Hom}(X,Y)\to\operatorname{Hom}(X,Z)\\ (g\circ f)(x) = g(f(x))$$

where $\operatorname{Hom}(A,B)$ is the set of functions from $A$ to $B$ (this can easily be extended to situations when the codomain of $f$ doesn't equal the domain of $g$, but the image of $f$ is contained in the domain of $g$). Notice that this isn't a binary operation unless $X=Y=Z$ and hence is not "multiplication".

However, you correctly notice that if we restrict function composition to $$\operatorname{Sym}(X) = \{f\colon X\to X\mid f\ \text{is invertible}\}$$ we indeed get group structure with $\circ$ as multiplication, identity function $\operatorname{id}_X$ as identity and function inverse $f\mapsto f^{-1}$ as inverse. In case when $X$ is finite set and $|X| = n$, we usually denote $\operatorname{Sym}(X)$ with $S_n$ and call it group of permutations.

When you define group homomorphisms, you do use composition of functions in the definition, but perhaps in not so obvious manner, if you haven't seen it before. Let $G$ and $H$ be groups with multiplications $\mu_G\colon G\times G\to G$ and $\mu_H\colon H\times H\to H$. Then, $f\colon G\to H$ is group homomorphism if $$f\circ\mu_G = \mu_H\circ(f\times f)$$ where $f\times f\colon (x,y)\mapsto (f(x),f(y))$.

Let me elaborate on that. If we denote $\mu_G(x,y) = x\star y$ and $\mu_H(x,y) = x\cdot y$, then $$(f\circ\mu_G)(x,y) = f(\mu_G(x,y)) =f(x\star y),$$ while $$(\mu_H\circ(f\times f))(x,y) = \mu_H(f(x),f(y)) = f(x)\cdot f(y).$$