Let $G$ be a group such that for all $x,y \in G, (xy)^3 = x^3y^3$ and there's no element with order $3$. Show that $xy^2=y^2x$ and $G$ is abelian.

Question

Let $G$ be a group such that for all $x,y \in G, (xy)^3 = x^3y^3$ and there's no element with order $3$. Show that $xy^2=y^2x$ and $G$ is abelian.

Any idea or some hints? Thanks in advanced.

Answer 1

$G$ is finite from the tags, though it is not stated.

From $(ab)^3 = a^3b^3$, we get that $(ab)^2 = (ab)^3(ab)^{-1} = a^3b^3b^{-1}a^{-1} = a^3b^2a^{-1}$. Thus, $abab = a^3b^2a^{-1}$ or $(ba)^2 = a^2b^2$.

So $b^3a^3= (ba)^3 = (ba)(ba)^2 = ba^3b^2$. Therefore, $b^2a^3 = a^3b^2$, so every cube commutes with every square.

Let $g\in G$. Since $|g|$ is not divisible by $3$, $|g^3| = |g|/\gcd(|g|,3) = |g|$, so $\langle g^3\rangle = \langle g\rangle$. Therefore, there exists $k\gt 0$ such that $g=(g^3)^k = (g^k)^3$. In particular, every element of $G$ is a cube.

This means that every element of $G$ commutes with ever square, so $G^2\subseteq Z(G)$.

But now we have that $a^3b^3 = aa^2b^2b = ab^2a^2b$, so $a^2b^2=b^2a^2 = (ab)^2$. And it is well known that from $(ab)^2=a^2b^2$ we may conclude that $G$ is abelian.

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As stated in the sci.math thread I linked to in the comments, a variant that does not require the assumption of finiteness is:

Let $G$ be a group with no $3$-torsion in which $(ab)^3 = a^3b^3$ for all $a,b\in G$. Then $G$ is abelian.

This is the problem as stated, if we don’t look at the tags to find out we are supposed to assume $G$ is finite. This statement is also correct.

Proof: as above, we know that every cube commutes with every square. Hence: $$\begin{align*} [a,b]^3 &= (a^{-1}b^{-1}ab)^3\\ &= a^{-3}b^{-3}a^3b^3\\ &= a^{-3}b^{-3}b^2a^3b\\ &= a^{-3}b^{-1}a^3b\\ &= [a^3,b]. \end{align*}$$ In particular, $[a,b^2]^3= [a^3,b^2] = 1$ (because squares and cubes commute). Since $G$ has no $3$-torsion by assumption, this means that $[a,b^2]=1$, so squares are central in $G$.

Now we can proceed as before: $(ab)^3 = a^3b^3 = ab^2a^2b$, so $(ba)^2 = b^2a^2$ and we conclude that $G$ as abelian as before.

Answer 2

Here is a hint to get you started:

$(xy)^3 = x^3y^3 \iff y(xy)x = yxyx = xxyy = x(xy)y$

You need to prove that $xyy=yyx$ and then use that to prove $xy=yx$.

Answer 3

Let $x,y \in G$. First, we'll prove that $xy^2 = y^2x$. From $(xy)^3 = x^3y^3$, \begin{align*} xyxyxy &= xxxyyy \\ yxyx &= xxyy \\ \end{align*} we obtain $(yx)^2 = x^2y^2 \dots (1)$. Next, consider that \begin{align*} (xy)^9 &= x^9y^9 \\ (yx)^8 &= x^8y^8 \\ (x^2y^2)^4 &= x^8y^8 \ (from \ (1)) \\ xxyyxxyyxxyyxxyy &= xxxxxxxxyyyyyyyy \\ (y^2x^2)^3 &= (x^2)^3(y^2)^3 \\ (y^2)^3(x^2)^3 &= (x^2)^3(y^2)^3 \end{align*} Since there are no elements in $G$ with order $3$, we get $y^2x^2 = x^2y^2$. Now, consider $(1)$.

\begin{align*} yxyx &= xxyy \\ xyxyx &= xxxyy \\ xxyyx &= xxxyy \\ yyx &= xyy ... (2) \end{align*}

Hence, $xy^2 = y^2x$. Next, from $(2)$, we get

\begin{align*} xyy &= yyx \\ xxyy &= xyyx \\ yxyx &= xyyx \\ yxy &= xyy \\ yx &= xy \end{align*}

Hence, $xy=yx$, for all $x,y \in G$, which proves that $G$ is abelian.

Is it correct?