Groups

There is some new notation that we'll use temporarily:

Write $a\star b$ for $\mu (a,b)$. Then the properties above become:

1. $a\star (b\star c)=(a\star b)\star c$
2. $a\star e=a=e\star a$
3. $a\star b=e=b\star a$

Example

For instance, all the sets and their associated operations:

1. $\mathbb{Z},+$
2. $\mathbb{Q},+$
3. $\mathbb{R},+$
4. $\mathbb{C},+$
   Are all valid groups. That's because:
5. They are all associative (easy to show)
6. The identity is $0$
7. The inverse is $-a$ for each $a\in G$ since $a+(-a)=0=(-a)+a$

Another example is all the sets and their associated operations:

1. $\mathbb{Z},\cdot$
2. $\mathbb{Q},\cdot$
3. $\mathbb{R},\cdot$
4. $\mathbb{C},\cdot$
   None of these are valid groups. That's because:
5. They are all associative
6. The identity is $1$ for all of them
7. $0$ doesn't have an inverse here!
   What we do here is just "throw away" the dud which is $0$. If we only consider those sets with $0$ removed then:
8. $(\mathbb{Q}-\{0\},\cdot )$ is not a group where each $\frac{a}{b}$ has its inverse $\frac{b}{a}$
9. The other groups follow a similar patter (other than $(\mathbb{Z}-\{0\},\cdot )$ obviously)

Another example is the set of functions from $\mathbb{R}\to \mathbb{R}$ and the binary operation $+$ is just adding their outputs. Namely $(\text{Fun}(\mathbb{R},\mathbb{R}),+)$ where:
$(f+g)(x):=f(x)+g(x)$

1. Associativity comes from associativity of $\mathbb{R},+$
2. The identity is the zero function $o(x)=0$
3. The inverse of some $f(x)$ is just $-f(x)$ since $f(x)+(-f(x))=o(x)=(-f(x))+f(x)$

Another example is doing compositions, namely $(\text{Fun}(\mathbb{R},\mathbb{R}),\circ )$ where for all $f,g$ in this set:
$(f\circ g)(x)=f(g(x))$
Here:

1. Associativity is definitely here
2. Identity is just $I(x)=x$
3. The inverse is the problem! There isn't always an inverse. Namely $f$ has an inverse iff $\mathrm{\exists }g$ where $(f\circ g)=I=(g\circ f)$.