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This problem is from abstract algebra and I am trying to solve but I am completely lost. The problem is as follows:

If $f\in S_3$, show that $f^6 = I$.

This section has to do with $A(S)$, the set of $1$ to $1$ mappings of $S$ onto itself. The textbook's answer is $S=\{x_1,x_2,x_3\}$ and $f\in S_3$ and $f(x_1)\neq x_1$. Since there are only a finite number of choices in $S$, either $f^2(x_1) = x_1$ or $f^3(x_1) = x_1$. This is similar for $x_2$ and $x_3$, $f$ satisfies $f^2(x)=x$ or $f^3(x)=x$. This gives $f^6(x)=x$ for all members of $S$.

I don't understand this answer could someone please elaborate?

HallaSurvivor
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Lavorizia Vaughn
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1 Answers1

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Say $f \in S_3$.

Then $f$'s reason in life is to permute a three element set. That is $f$ should be some bijection from $\{x_1, x_2, x_3\}$ to itself.

For instance, we might have $f(x_1) = x_1$, $f(x_2) = x_3$, and $f(x_3) = x_2$.

If this isn't clear you should spend some time rereading whatever part of your book or notes introduced the symmetric groups.

Now, what does it mean for $f^6 = I$? If this is what we're trying to prove, we had better understand the statement!

Well $I$ is the identity mapping that just fixes all the elements. So $I(x_1) = x_1$, $I(x_2) = x_2$ and $I(x_3) = x_3$.

Next, $f^n$ is the map we get by applying $f$ $n$-many times. So for instance, $f^3(x_2) = f(f(f(x_2)))$.

Now, you should take a moment and check this by hand for at least two examples.

  1. If $f$ fixes $x_1$ and swaps $x_2$ and $x_3$, manually compute $f^6(x_1)$, $f^6(x_2)$, and $f^6(x_3)$. Check that in each case $f^6(x) = I(x) = x$

  2. If $f(x_1) = x_2$, $f(x_2) = x_3$, and $f(x_3) = x_1$, do the same thing. Manually compute $f^6$ on each input and verify that it agrees with $I$.

At this point, since there's only $6$ elements of $S_3$, you could (in theory) just check all of them! But what if somebody asked you to prove a similar theorem about $S_{100,000}$? Is there a more clever approach that saves you having to check all the cases?

The answer, of course, is yes! The idea is to figure out a power high enough to guarantee that $f^n(x) = I(x) = x$ for each $x_1$, $x_2,$ and $x_3$.

Let's look at $x_1$ -- all the other cases are the same.

If $f(x_1) = x_1$, then we're done.

If $f(x_1) = y \neq x_1$, then we only have two options. Either $f(y) = x_1$, or $f(y) = z$.

$\quad \quad$ If $f(y) = x_1$, then we see $f^2(x_1) = f(f(x_1)) f(y) = x_1$, and we're done.

$\quad \quad$ If $f(y) = z$, then since $f$ is a bijection, we must have $f(z) = x_1$ (do you see why?). In this case, $f^3(x_1) = x_1$.

So we know that either $f$, $f^2$, or $f^3$ fixes $x_1$.

But if $f^2(x_1) = x_1$, then $f^6(x_1) = f^2(f^2(f^2(x_1))) = x_1$ too. Similarly, if $f^3(x_1) = x_1$, then $f^6(x_1) = f^3(f^3(x_1)) = x_1$!

So no matter what, we see that $f^6(x_1) = x_1$!

Of course, nothing was special about $x_1$ here, and the exact same argument shows that $f^6(x_2) = x_2$ and $f^6(x_3) = x_3$.

So for all inputs, $f^6(x_i) = x_i = I(x_i)$, so $f^6 = I$. Exactly what we set out to prove.

As an aside, this can be generalized quite far! In fact, for any finite group $G$, $x^{|G|} = 1$. See here, for instance.

I hope this helps ^_^

HallaSurvivor
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