Is it true that $O(ab)=O(ba)$, where $G$ is a group and $a,b \in G$?

Question

Is it true that $O(ab)=O(ba)$, where $G$ is a group and $a,b \in G$?

Suppose $O(a)$ and $O(b)$ are finite and also $O(ab)$ and $O(ba)$ are finite. Then $\operatorname{lcm} (|a|,|b|) = \operatorname{lcm}(|b|,|a|)$. (Is that correct?)

Suppose $O(a)$ and $O(b)$ is finite but $O(ab)$ is infinite, how to prove that $O(ab) = O(ba)$?

If not give elements $a,b$ in a group $G$, such that $O(ab)$ is infinite but $O(ba)$ is finite.

Answer 1

(1) Conjugate elements have the same order, i.e. $\;\mathcal o(a)=\mathcal o(g^{-1}ag)\;$

(2) $\;ab=b^{-1}(ba)b\;$

Answer 2

Just to be clear, the order of $ab$ isn't necessarily the LCM of the orders of $a$ and $b$. For example, consider the cycles $a=(1,2)$ and $b=(1,2,3)$ in $S_3$: the LCM of their orders is 6, but $ab=(1,3)$ (or $(2,3)$ if right-to-left) has order 2.

However order is invariant under isomorphisms. More precisely, if $\phi: G \rightarrow H$ is a group isomorphism and $g \in G$, the order $|\langle g \rangle|$ of $g$ in $G$ is the same as $|\langle \phi(g) \rangle|$, the order of $\phi(g)$ in $H$.

In particular, conjugation by $b$ is an automorphism of $G$: define $\phi: G \rightarrow G$ by $\phi(g) = b^{-1} g b$. Then $ab = \phi(ba)$ so the two have the same order (whether finite or infinite).

Answer 3

Lemma: If the order of $ab$ is infinite, then the order of $ba$ is infinite.

Proof. Contrary to our claim, let $o(ba)= m$. So, $$\underbrace{(ba)(ba)(ba)\cdots (ba)}_{m}=e.$$ Then, $$\underbrace{(ab)(ab)\cdots (ab)}_{m-1}= b^{-1}a^{-1}=(ab)^{-1}.$$ Hence, $o(ab)\mid m$, which is a contrast.

So, you don't need to have the orders of $a$ and $b$ are finite.

Answer 4

Suppose $(ab)^k=1$. Then $(ab)^k aa^{-1} = a (ba)^k a^{-1}=1$. Thus, $(ba)^k=1$.

Answer 5

If $a$, $b$ commute, and the orders of $a$, $b$ are relatively prime, then the order of $ab$ is $O(ab)=O(a)\cdot O(b)$.

Answer 6

There is no need for assumptions on finiteness.

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Let $a,b\in G$ for a group $G$ and let $m\in\Bbb N$.

We have

$$\begin{align} b(ab)^mb^{-1}&=b\underbrace{(ab)\dots(ab)}_{m\text{ times}}b^{-1}\\ &=\underbrace{(ba)\dots(ba)}_{m\text{ times}}(bb^{-1})\\ &=(ba)^m. \end{align}$$

Therefore, if $(ab)^m=e$, then

$$\begin{align} e&=bb^{-1}\\ &=beb^{-1}\\ &=b(ab)^mb^{-1}\\ &=(ba)^m. \end{align}$$

It follows that $|ab|=|ba|$ (because $$(ab)^m=e\implies (ba)^m=e,$$ which gives $$(ab)^m=e\iff (ba)^m=e$$ by the symmetry of the argument; but $m$ was arbitrary; so, by looking at divisibility, their orders must be the same, which includes the case of infinite order, due to the contrapositive $$(ab)^k\neq e\iff (ba)^k\neq e$$ following from the above for infinite order for each $k\in\Bbb N$.)

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Isomorphisms between groups preserve orders of elements. Conjugation

$$\begin{align} c_g:G&\to G,\\ x&\mapsto gxg^{-1} \end{align}$$

by any element $g\in G$ is an isomorphism. We have

$$\begin{align} c_b(ab)&=b(ab)b^{-1}\\ &=(ba)(bb^{-1})\\ &=(ba)e\\ &=ba. \end{align}$$

Therefore, $|ab|=|ba|$.

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We have $b(ab)^1b^{-1}=(ba)^1$. Consider this an $m=1$ base case for

$$b(ab)^mb^{-1}=(ba)^m.$$

Suppose

$$b(ab)^kb^{-1}=(ba)^k\tag{1}$$

for some $k\in\Bbb N$.

Consider when $m=k+1$. We have

$$\begin{align} b(ab)^{k+1}b^{-1}&=b(ab)(ab)^kb^{-1}\\ &=(ba)\color{red}{(b(ab)^kb^{-1})}\\ &\stackrel{(1)}{=}(ba)\color{red}{(ba)^k}\\ &=(ba)^{k+1}, \end{align}$$

which is $(1)$ with $k$ replaced by $k+1$. Therefore, by induction on $m$, we have

$$b(ab)^mb^{-1}=(ba)^m.$$

As before, this gives $|ab|=|ba|$.