Why Do $xy$ and $yx$ Have the Same Order in a Group?

For any elements $x,y$ in a group $G$, the products $xy$ and $yx$ have the same order. The key is that they are conjugate: one is obtained from the other by conjugating with a group element. This proof does not require $G$ to be finite or commutative.

The proof

Recall that the order $o(g)$ of an element $g$ is the least positive integer $n$ for which $g^n=e$, if such an integer exists. If no positive power of $g$ equals the identity, $g$ has infinite order.

Conjugate $xy$ by $x$:

$$x^{-1}(xy)x=(x^{-1}x)yx=yx.$$

So $yx$ is a conjugate of $xy$. More explicitly, for every positive integer $n$,

$$(yx)^n=\bigl(x^{-1}(xy)x\bigr)^n=x^{-1}(xy)^n x.$$

The middle factors cancel when the product is expanded. Consequently,

$$(yx)^n=e \quad\Longleftrightarrow\quad x^{-1}(xy)^n x=e \quad\Longleftrightarrow\quad (xy)^n=e.$$

The last equivalence follows by multiplying by $x$ on the left and $x^{-1}$ on the right. Thus $xy$ and $yx$ have exactly the same positive powers that equal the identity. If there is a least such exponent, it is the same for both elements; if there is none, both have infinite order. Therefore,

$$o(xy)=o(yx).$$

A worked example in $S_3$

Take $G=S_3$, the group of permutations of $\{1,2,3\}$, and let $x=(12)$ and $y=(23)$. Use the standard convention that permutations are composed right to left. Then

$$xy=(12)(23)=(123), \qquad yx=(23)(12)=(132).$$

These are different permutations, but each is a 3-cycle, so each has order 3. The conjugation calculation explains why their orders agree: $x^{-1}=x$ here, and $x^{-1}(xy)x=yx$. The general result says more than this example: the products need not be equal, but their orders always agree.

Why conjugation preserves order

More generally, for any $g,h\in G$, the elements $g$ and $h^{-1}gh$ have the same order. For every positive integer $n$,

$$(h^{-1}gh)^n=h^{-1}g^n h.$$

Hence $(h^{-1}gh)^n=e$ exactly when $g^n=e$. This is why conjugation preserves both finite order and infinite order. In the question, take $g=xy$ and $h=x$.

You can also verify the conjugacy in the other direction: $y(xy)y^{-1}=yx$. Either identity identifies the same pair as conjugate elements.

Common mistakes

• Assuming $xy=yx$. That equality is true in an abelian group, but not generally. The claim concerns the orders, not the elements themselves.
• Trying to reorder the factors. You do not need commutativity. The cancellation in $x^{-1}(xy)x=yx$ uses associativity and the inverse law only.
• Using only finite-order reasoning. If the order is infinite, compare the identity powers instead: the equivalence $(xy)^n=e$ if and only if $(yx)^n=e$ holds for every positive integer $n$.
• Claiming the converse for arbitrary pairs. This argument shows that $xy$ and $yx$ are conjugate, so they have the same order. Having the same order alone does not generally imply that two elements are conjugate.

The same reasoning is useful whenever a problem replaces an element by a conjugate: properties determined by its powers, such as its order, are unchanged.