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Problem statement: if $G$ is a group and $a,b\in G$, prove that if $|ab|=k$, then $|ba|=k$.

I'm going to assume one thing: that $|a|$ means $|\langle a \rangle|$, that is, the size of the subgroup generated by $a$.

I argued that the size of each subgroup is less than or equal to the size of the other subgroup, and that therefore, their sizes are equal.

($|ba|\leq |ab|$) Every element $g\in \langle ba \rangle$ has the form $(ba)^n$, where $n\in \mathbb{Z}$. This element can be generated by $n+1$ compositions of $ab$: $(ab)^{n+1}= a(ba)^nb$. This means that every element in $\langle ba\rangle$ can be generated in $\langle ab\rangle$ and that therefore $\langle ab\rangle$ must have at least as many elements as $\langle ba\rangle$.

I used the same reasoning to conclude $|ab|\leq |ba|$.

What is wrong with this reasoning (if at all)?

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    Even more is true: $ab$ and $ba$ are always conjugate, thus they have the same order.2012-10-17

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Another way, perhaps shorter. First, prove the easy

Lemma: For any elements $\,x\,,\,g\,$ in a group $\,G\,$ , we have that $\,|x|=|x^g|=|g^{-1}xg|\,$

Well, we're then done since $\,ba=(ab)^a\,$

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    Oh, I have never seen it before. thanks for explaining.2012-10-17
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As Thomas says, it is false that the elements in $\langle ab \rangle$ are necessarily in $\langle ba \rangle$, but your argument can be recovered :

Consider the function $f : \langle ab \rangle \to \langle ba \rangle$ defined by $f(g) = bga$. It is well-defined because any element $g \in \langle ab \rangle$ can be written $(ab)^k$, and then we see that $b(ab)^ka = (ab)^{k+1}$ is in $\langle ba \rangle$, as you noted. Since we're in a group, $f$ is injective : if $f(g) = f(h)$, you can cancel $a$ and $b$ to obtain $g=h$. This shows that $| \langle ab \rangle| \ge |\langle ba \rangle|$.