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Proposition: Let $G$ be a group and $a,b \in G$. Show that $|ab| = |ba|$.

Proposed Proof: Assume that $|ab| = n<\infty$. This tells us that $(ab)^n = b^na^n = 1$, from which we conclude that $b^n = (a^n)^{-1}$. Using this we can conclude that $$(ba)^n = a^nb^n = a^n(a^n)^{-1} = 1.$$ So again it remains to show that it is of lowest order. Assume for contradiction that there exists some $s \in \mathbb{N}$ with $s<n$ for which $(ba)^s = 1$. Then $a^sb^s = 1$, implying that $b^s = (a^s)^{-1}$. If we then consider $(ab)^s$, we get that $$ (ab)^s = b^sa^s = (a^s)^{-1}a^s = 1, $$ which contradicts the minimality of $n$ as $|ab| = n$. Therefore $|ba| = n$ as required.

Questions: Is this the best way to show the proof? Specifically for minimality is there a better way to do this without resorting to a contradiction? If I remember from algebra courses I've taken it feels like a standard approach but it's been a while. Any advice or fixes if the proof is wrong is appreciated!

Irving Rabin
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    Why is $(ab)^n = b^n a^n$? – Jose Avilez May 05 '22 at 02:19
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    The equation $(ab)^n=b^na^n$ is generally not true, even if $n=|ab|$. You should use the fact $ab$ and $ba$ are conjugate, which implies if a power of one is trivial then the same power of the other is too. – anon May 05 '22 at 02:25
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    @JoseAvilez I thought that was actually standard, woops. So it's only true when $a,b$ are commuting elements of $G$? – Irving Rabin May 05 '22 at 02:40
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    For example, if $G=S_3$, $a=(12)$, $b=(23)$. Then $ab=(132)$ and $|ab|=3$. However, $e=(ab)^3\neq b^3a^3=ba=(123)$. – kabenyuk May 05 '22 at 02:43
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    Show that $|gxg^{-1}| = |x|$ for every $x$ and $g$ in $G$. Next note that $ab = a(ba)a^{-1}$. – azif00 May 05 '22 at 03:56
  • For any commuting elements $a$ and $b$ and exponent $n$ we have $(ab)^n=a^nb^n$, not to mention in this case any powers of $a$ and $b$ commute too, i.e. $a^kb^\ell=b^\ell a^k$ for any integer exponents $k$ and $\ell$. For the specific case of $n=-1$ there is the socks-and-shoes rule $(xy)^{-1}=y^{-1}x^{-1}$ which holds for any group elements $x$ and $y$. – anon May 06 '22 at 19:05

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