# Prove that $\Gamma\left(-a\right)=\left[\Gamma\left(a\right)\right]^{-1}$ for $\Gamma:\mathbb{Z}\rightarrow \mathcal{B}\left(A,A\right)$

I am working through various problems in Bloch's Proofs and Fundamentals and I'm stuck on this problem (in need of hints):

Let $A$ be a set. A $\mathbb{Z}$-action on $A$ is a function $\Gamma:\mathbb{Z}\rightarrow \mathcal{B}\left(A,A\right)$ (the set of all bijective functions from $A\rightarrow A$) that satisfies the following properties: $\Gamma\left(0\right)=1_A$ and $\Gamma\left(a+b\right)=\Gamma\left(a\right)\circ\Gamma\left(b\right)$ for all $a,b\in\mathbb{Z}$.

Prove that $\Gamma\left(-a\right)=\left[\Gamma\left(a\right)\right]^{-1}$.

Questions: What do they mean by "$\Gamma\left(a\right)\circ\Gamma\left(b\right)$"? This looks like the composition of two functions, but they are acting upon two different elements $a$,$b$? Also, I am assuming $\left[\Gamma\left(a\right)\right]^{-1}=\frac{1}{\Gamma\left(a\right)}$? If it had meant the inverse I would assume it would be written simply as "$\Gamma^{-1}\left(a\right)$".

Now suppose the simplest case, $a=0$. Then $a=b=0$ and $\Gamma\left(-0\right)=\Gamma\left(0\right)=\Gamma\left(0+0\right)=\Gamma\left(0\right)\circ\Gamma\left(0\right)$ (and I don't know how to visualize the "$\circ$" operation between them, as I've previously stated). But now $\Gamma\left(0\right)\circ\Gamma\left(0\right)=I_A\circ I_A=I_A$. Does $I_A=1/I_A$?

• You might want to change the title: lotta people would think you're referring to the Gamma function $\Gamma(s)=\int_0^\infty e^{-t}\,t^s\,dt/t$. – paul garrett May 17 '15 at 17:58
• @paulgarrett I thought that may have been the case. I've edited accordingly. – jm324354 May 17 '15 at 18:00
• $\Gamma(a) \in \mathcal B(A,A)$, so $\left[\Gamma(a)\right]^{-1}$ refers to the map $f \in A$ such that $\Gamma(a) \circ f = 1_A$. This is different then $\Gamma^{-1}$, which is a function that takes members of $\mathcal B(A,A)$ (which are themselves maps) and maps them to integers. – dalastboss May 17 '15 at 18:24

• What does $\Gamma(a) \circ \Gamma(b)$ mean?:

$\Gamma$ takes integers and maps them to functions. This is often a point of confusion for people so it bares repeating: $\Gamma$ is a function which outputs functions. Specifically, $$x \in \mathbb Z \implies \Gamma(x) \in \mathcal B(A,A)$$ That is what $\Gamma : \mathbb Z \rightarrow \mathcal B(A,A)$ means. Then since $\Gamma(x) \in \mathcal B(A,A)$, $\Gamma(x) : A \rightarrow A$. If $y \in \mathbb Z$ as well, then the same can be said for $\Gamma(y)$. Thus since they are both maps from $A$ to $A$, they can be composed: write $f = \Gamma(x)$ and $g = \Gamma(y)$. Then for $e \in A$, $$(\Gamma(x) \circ \Gamma(y))(e) = (f \circ g)(e) = f(g(e)) \in A$$

• Does $\left[\Gamma(a)\right]^{-1} = \frac{1}{\Gamma(a)}$?:

It is a matter of notation. Whenever you have a binary operation (for example, $\star$) with an identity (for example, $I_\star$), the inverse of an element $a$, denoted $a^{-1}$, is the element such that $a \star a^{-1} = I_\star$. So in this problem, your binary operation is $\circ$, and the identity $I_A \in \mathcal B(A,A)$ is the is the identity function (i.e. $\forall x \in A. \; I_A(x) = x$), so since $\Gamma(a) \in \mathcal B(A,A)$, then $\left[\Gamma(a)\right]^{-1}$ refers to the map $f \in A$ such that $\Gamma(a) \circ f = I_A$. (Side note: the identity is often called $1_A$ instead of $I_A$. They mean the same thing.)

That is what is meant by $\left[\cdot\right]^{-1}$. Since $\Gamma(a)$ is a function, it would be considered abuse of notation to write $\frac{1}{\Gamma(a)}$. This would be like saying if $f(x) = 2x$, then $\frac{1}{f}(x) = \frac{1}{2}x$ because $\frac{1}{2}x$ is the inverse function of $2x$.

• Does $I_A = \frac{1}{I_A}$?:

As explained above, $\frac{1}{I_A}$ is abuse of notation. However, it is true that $I_A = I_A^{-1}$. The identity element of a given group is always its own inverse.

• $\left[\Gamma(a)\right]^{-1}$ versus $\Gamma^{-1}(a)$:

$\Gamma(a) \in \mathcal B(A,A)$, so $\left[\Gamma(a)\right]^{-1}$ refers to the map $f \in A$ such that $\Gamma(a) \circ f = 1_A$. This is different then $\Gamma^{-1}$, which is a function that takes members of $\mathcal B(A,A)$ (which are themselves maps) and maps them to integers. Therefore, in order for the expression "$\Gamma^{-1}(a)$" to make sense, $a$ would actually have to be a function from the set $\mathcal B(A,A)$ rather than an integer, where as in $\left[\Gamma(a)\right]^{-1}$, $a$ is just an integer.

Yes, the notation is confusing. It can be difficult to get used to. I recommend keeping all definitions handy when learning new material, and please comment if you have further confusion.

$\Gamma(a)$ and $\Gamma(b)$ are elements of $\mathcal B(A,A)$, hence both are maps $A\to A$ and can be composed via $\circ$.

Also $\Gamma(a)$ is a bijection $A\to A$ so that it makes sens to speak of the inverse map, which is as usually denoted $[\Gamma(a)]^{-1}$. (It cannot be $\frac1{\Gamma(a)}$ simply because there is no structure involved here that would allow division)

On the other hand $\Gamma^{-1}(a)$ would denote an element $n\in \mathbb Z$ such that $\Gamma(n)=a$, so this would require $a\in\mathcal B(A,A)$, ot $a\in\mathbb Z$.

Note that $\Gamma(a)$ is a member of $\mathcal B (A,A)$, i.e. it is a bijective function from $A$ to $A$. The same holds for $\Gamma(b)$. Therefore, it does make sense to compose them and speak of $\Gamma(a) \circ \Gamma(b)$. Similarly, given that $\Gamma(a)$ is bijective, it makes sense to speak about its inverse denoted $\Gamma(a)^{-1}$.