# Proving the following set is a group

I've been trying to solve a question related to group theory. If we consider that the question says that $$f^0$$ is the identity and $$f^0$$ and $$f$$ is the same and just an error, then how identity $$\times$$ identity = another function.

I can easily prove that it is associative and maintains closure property using cayley table but I can't understand how $$f^0$$ is both an identity function as well as gives some other value when multiplied by itself.

• $f^0$ is not $f$, it is the identity map from $X$ to $X$ Commented Feb 25, 2019 at 11:34
• Why do you say that $f^0$ gives another value when multiplied by itself?
– Javi
Commented Feb 25, 2019 at 11:34
• What is the difference between identity and identity map. Also I considered f0=f Commented Feb 25, 2019 at 12:13
• In this case the identity map is the identity of the group. If you consider $f^0$ to be $f$ then it is not the identity, the question doesn't do that. Commented Feb 25, 2019 at 12:16
• What does f refer to then ? Commented Feb 25, 2019 at 12:17

We define $$f^2$$ as being the same function as $$f \circ f$$, so

$$f^2(x) = (f \circ f)(x) = f(f(x))$$

And then we can extend the notation:

$$f^3(x) = (f^2 \circ f)(x) = f(f(f(x))) \\ f^4(x) = (f^3 \circ f)(x) = f(f(f(f(x))))$$

and so on. And it also makes sense to say that $$f^1(x) = f(x)$$.

But what about $$f^0(x)$$ ? Well, if $$f^0$$ exists then we want it to satisfy the following identity:

$$f(x) = f^1(x) = (f^0 \circ f)(x) = f^0(f(x))$$

So $$f^0$$ is the identity function i.e. $$f^0(f(x))=f(x)$$ and $$f^0(x)=x$$. And then if we compose $$f^0$$ with itself we have

$$(f^0)^2(x)=(f^0 \circ f^0)(x) = f^0(f^0(x))) = f^0(x)$$

so $$f^0 \circ f^0 = f^0$$. And also

$$(f^0 \circ f^n)(x) = f^0(f^n(x)) = f^n(x)$$

In your example, $$f$$ maps $$1$$ to $$2$$ and $$2$$ to $$3$$, so $$f^2(1)=f(f(1))=f(2)=3$$. Similarly $$f^2(2) = f(f(2)) = f(3) = 4$$. Continuing like this, we have

$$f^2=\{(1,3),(2,4),(3,1),(4,2)\}$$

In the same way, you can work out what $$f^3$$ is, and eventually you will find that $$f^4(x)=x$$, so $$f^4=f^0$$.

You should now be able to show that $$F$$ with the operation of composition of functions satisfies all four properties of a group.

• To prove closure and associative we can show that the elements resulting are from the set of f. What would be the inverse of this. Commented Feb 25, 2019 at 14:35
• Once you know that $f^0$ is the identity for composition of functions and $f^4=f^0$ then you can re-write this as $f^3 \circ f = f^0$. What does this suggest should be the inverse of $f$ ? Commented Feb 25, 2019 at 14:45
• f^3 yes. So we only need to find the inverse of f ? Commented Feb 25, 2019 at 14:54
• You need to show that all four members of $F$ have an inverse. But you know that $f^0$ is its own inverse, and since $f^3$ is the inverse of $f$ then you also know that $f$ is the inverse of $f^3$. So you are almost there. Commented Feb 25, 2019 at 14:58
• also , 1 last query, f and f1 are both the same ? Commented Feb 25, 2019 at 15:09