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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, $x = e$.

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    $\begingroup$ 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 '17 at 23:08
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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 $x = e$.

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  • $\begingroup$ 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 '17 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 '17 at 23:19
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    $\begingroup$ 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 '17 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 '17 at 23:27
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    $\begingroup$ 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 '17 at 23:37
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$$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$

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.

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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.

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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 $$x=x^{um+vn}=(x^m)^u(x^n)^v=e^ue^v=e.$$

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