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I just want to make sure I'm alright on this proof and whether the contradiction is necessary. In some of the proofs I've checked here they simply show that $x^n = 1$ to conclude that $|x| = n$, but shouldn't we also need to show that it's the minimal such integer as I've tried to do below?

Prop: Let $x \in G$, show that $|x| = |x^{-1}|$.

Proof: Assume that $|x| = n < \infty$, then we have that $(x^{-1})^n = (x)^{-n} = (x^n)^{-1} = 1$. Therefore it remains to show that $n$ is the minimal such integer. Assume for contradiction that $\exists \; r \in \mathbb{Z}^+$ such that $(x^{-1})^r = 1$ with $r < n$. Then by the similar logic to before $$ \begin{align} (x^{-1})^r &= 1 \\ x^r\cdot(x^r)^{-1} &= x^r \cdot1 \\ 1 &= x^r. \end{align} $$ But this contradicts the minimality of $n$ as $|x| = n$. If $|x| = \infty$, then assume for contradiction that $|x|^{-1} = k < \infty$. Identically to the above we can arrive at $$ \begin{align} (x^{-1})^k &= 1 \\ x^k\cdot(x^k)^{-1} &= x^k \cdot1 \\ 1 &= x^k. \end{align} $$ which contradicts that $x$ is of infinite order.

Is it unnecessary to use the contradictions, if so what's a hint at the direct proof?

RobPratt
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Irving Rabin
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    Looks elegant enough to me – gt6989b May 31 '22 at 17:45
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    It's fine. Another approach is from $(x^{-1})^n = (x^n)^{-1}$ you can conclude $|x| \le |x^{-1}|$. And then swap the roles of $x$ and $x^{-1}$ to get equality. – ErikR May 31 '22 at 17:58
  • Looks cool to me – Riemann'sPointyNose May 31 '22 at 18:34
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    Another approach (not by contradiction) for the case when $x$ has finite order. Let $|x|=n$. Then, since inverses are unique, $x^{-1} = x^{n-1}$. Note that $x^{k} = 1$ iff $k|n$ and $(n-1)$ and $n$ are coprime, so we must have $n \mid |x^{n-1}|$. Furthermore, $n$ suffices, so we find that $|x^{n-1}| = |x^{-1}| = n$. – legionwhale May 31 '22 at 19:01
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    For a direct proof, see for example this duplicate. – Dietrich Burde May 31 '22 at 19:10
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    Note that $a^k=e\iff (a^{-1})^k = e$, hence the set (actually an ideal) ${n\in\mathbb{Z}\mid a^n=e}$ is exactly the same as the set ${n\in\mathbb{Z}\mid (a^{-1})^n = e}$. Since the order of $a$ is the nonnegative generator of this ideal, we conclude that $a$ and $a^{-1}$ have the same order. Alternatively, if the order is the cardinality of $\langle a\rangle$, note that $a\in\langle a^{-1}\rangle$ and $a^{-1}\in\langle a\rangle$, hence $\langle a \rangle = \langle a^{-1}\rangle$. – Arturo Magidin May 31 '22 at 19:18

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