7
$\begingroup$

Full Problem: Prove that for any element $x$ in a group $G$ that satisfies $$x^9 = e \\ x^{11} = e,$$ where $e$ is the identity element, that $x$ itself must be $e$.

Is this as simple as showing that

  • $x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
  • $x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
  • $x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
  • $x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
  • $x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$

Therefore, $x = e$.

$\endgroup$
1
  • 4
    $\begingroup$ More generally, if $n$ and $m$ are any two relatively prime integers, then Bezout's lemma there exist $a$ and $b$ such that $an+bm=1$. In this case, if $x^n=e$ and $x^m=e$, then $x^{an+bm}$ is both $x^1$ and $e$. $\endgroup$
    – arctic tern
    Mar 18, 2017 at 23:08

4 Answers 4

Reset to default
13
$\begingroup$

Your reasoning is correct but here is a more direct argument.

Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$. Therefore, the order of $x$ is $1$ so $x = e$.

$\endgroup$
5
  • $\begingroup$ The proof of "$x^n=e\implies n$ is divisible by the order of $x$" is more or less a generalised version of this problem, so I wouldn't call it "more direct". $\endgroup$
    – Arthur
    Mar 18, 2017 at 23:16
  • $\begingroup$ @Arthur It avoids several steps in the original proof and gets at the essence of the problem. Sometimes a more general solution can be simpler and more direct. $\endgroup$
    – Qudit
    Mar 18, 2017 at 23:19
  • 2
    $\begingroup$ It doesn't avoid several steps, it hides them by quoting a theorem that uses basically the same steps in its proof. I'm not saying it's an invalid solution, but I do contest the "more direct" claim ("direct" is not the same as "short" or "elegant"), although that is a subjective thing. $\endgroup$
    – Arthur
    Mar 18, 2017 at 23:24
  • $\begingroup$ All it requires is knowing that if $x^n = e$ then the order of $x$ divides $n$ and that the gcd is $1$. The result in Bernard's answer is what would be a generalization (which my answer essentially proves) not dividing the order. Anyway, I am done with this pointless quibbling. $\endgroup$
    – Qudit
    Mar 18, 2017 at 23:27
  • 2
    $\begingroup$ I am not done. What you say is the whole point of my quibbling. You say "knowing that if $x^n=e$ then the order of $x$ divides $n$", and I say "what proof do you have of that?" and then, if you're in an accommodating mood, you will give me a proof. That proof will look almost identical to how the OP solved his problem, and you will go through exactly the same steps, just with general exponents instead of $9$ and $11$. That is what I mean when I say "It doesn't avoid steps, it hides them". $\endgroup$
    – Arthur
    Mar 18, 2017 at 23:37
8
$\begingroup$

$$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$

As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.

Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:

$$11=9\cdot 1 + 2\\ 9=2\cdot 1 + 7\\ 7=2\cdot 1 + 5\\ 5=2\cdot 1 + 3\\ 3=2\cdot 1 + 1$$

You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.

$\endgroup$
7
$\begingroup$

Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to $$e=x^9=x(x^2)^4=xe^4=x$$and you're done.

$\endgroup$
5
$\begingroup$

It can be as simple as what you proved, but it can be shorter. Actually you can prove the following result:

If in a group, $x^m=e$ and $x^n=e$ for coprime $m$ and $n$, then $x=e$.

Indeed we have a Bézout's relation: $\; um+vn=1$, so $$x=x^{um+vn}=(x^m)^u(x^n)^v=e^ue^v=e.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.