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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.

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  • $\begingroup$ $f^0$ is not $f$, it is the identity map from $X$ to $X$ $\endgroup$
    – Robert Chamberlain
    Commented Feb 25, 2019 at 11:34
  • $\begingroup$ Why do you say that $f^0$ gives another value when multiplied by itself? $\endgroup$
    – Javi
    Commented Feb 25, 2019 at 11:34
  • $\begingroup$ What is the difference between identity and identity map. Also I considered f0=f $\endgroup$
    – Naman Sood
    Commented Feb 25, 2019 at 12:13
  • $\begingroup$ 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. $\endgroup$
    – Robert Chamberlain
    Commented Feb 25, 2019 at 12:16
  • $\begingroup$ What does f refer to then ? $\endgroup$
    – Naman Sood
    Commented Feb 25, 2019 at 12:17

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Naman Sood
    Commented Feb 25, 2019 at 14:35
  • $\begingroup$ 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 $f$ ? $\endgroup$
    – gandalf61
    Commented Feb 25, 2019 at 14:45
  • $\begingroup$ f^3 yes. So we only need to find the inverse of f ? $\endgroup$
    – Naman Sood
    Commented Feb 25, 2019 at 14:54
  • $\begingroup$ 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. $\endgroup$
    – gandalf61
    Commented Feb 25, 2019 at 14:58
  • $\begingroup$ also , 1 last query, f and f1 are both the same ? $\endgroup$
    – Naman Sood
    Commented Feb 25, 2019 at 15:09

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