Prove that this is an abelian group 
$G$ is a group. $f:\:G\to G,\:a\mapsto a^3$ is a group monomorphism. Prove that $G$ is abelian.

My idea is that it's sufficient to prove that $(aba^{-1}b^{-1})^3=e$. So
$$
(aba^{-1}b^{-1})^3=a^3b^3a^{-3}b^{-3}
$$
and I am stuck. Or
$$
(aba^{-1}b^{-1})^3=a^3(ba^{-1}b^{-1})^3=a^3ba^{-3}b^{-1}
$$
but this kind of effort does not seem to be of any help.
 A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$This proof does not require the finiteness of $G$, as assumed in the (very nice!) answer to the previous post.
Let $a, b$ be arbitrary elements of $G$.
From
$$
a b a b a b = (a b)^{3} = a^{3} b^{3} = a a a b b b
$$
we obtain, multiplying by $a^{-1}$ on the left and by $b^{-1}$ on the right,
$$
(b a)^{2} = a^{2} b^{2}.
$$
Thus
$$
(b a)^{4} = ((b a)^{2})^{2} = (a^{2} b^{2})^{2} = b^{4} a^{4}.
$$
Rewrite as
$$
b a (b a)^{3} = b a b^{3} a^{3} = b^{4} a^{4}.
$$
Multiply by $b^{-1}$ on the left and $a^{-3}$ on the right to get
$$
a b^{3} = b^{3} a.
$$
So $\Set{ b^{3} : b \in G } \le Z(G)$. This finishes it if $G$ is finite, because then $f$ is surjective. 
But even without the finiteness assumption, we now have
$$
(a b)^{3} = a^{3} b^{3} = b^{3} a^{3} = (b a)^{3},
$$
and since $f$ is a monomorphism, $a b = ba$.
A: Here's my version (revised) . . .

Since the map $\:x\mapsto x^3$ is a monomorphism,
\begin{align*}
&\hspace{1pt}(ab)^3 = a^3b^3,\text{ for all }a,b \in G.\tag{1}\\[6pt]
&\hspace{1pt}\text{If }a,b \in G,\text{ and }a^3 = b^3,\text{ then }a = b.\tag{2}\\
\end{align*}
Let $a,b \in G.$
\begin{align*}
\text{Then}&&a^3b^3a^{-3}&=(aba^{-1})^3&&[\text{by} (1)]\\
&&&= (aba^{-1})(aba^{-1})(aba^{-1})&&\\
&&&= ab^3a^{-1}&&\\[8pt]
\text{But then}&&a^3b^3a^{-3} &= ab^3a^{-1}&&\\
\implies&& a^2b^3&=b^3a^2&&\\[8pt]
\text{So we have}&&a^2b^3&=b^3a^2,\,\text{ for all }a,b \in G.&&\tag{3}\\
&&&\text{(thus, squares commute}&&\\
&&&\text{ with cubes)}&&\\[8pt]
\text{Then}&&(a^2b)^3 &= a^6b^3&&[\text{by} (1)]\\
&&&= b^3a^6&&[\text{by} (3)]\\
&&&= (ba^2)^3&&[\text{by} (1)]\\[8pt]
\text{But then}&&(a^2b)^3 &= (ba^2)^3&&\\
\implies&& a^2b&=ba^2&&[\text{by} (2)]\\[8pt]
\text{So we have}&&a^2b&=ba^2,\,\text{ for all }a,b \in G.&&\tag{4}\\
&&&\text{(thus, squares commute}&&\\ 
&&&\text{ with everything)}&&\\[8pt]
\text{Then}&&(ab)^6 &=((ab)^2)^3&&\\
&&&= ((ab)(ab))^3&&\\
&&&= (ab)^3(ab)^3&&[\text{by} (1)]\\
&&&= a^3b^3a^3b^3&&[\text{by} (1)]\\
&&&=a(a^2)b^3a^3(b^2)b&&\\
&&&=ab^5a^5b&&[\text{by} (4)]\\[8pt]
\text{But then}&&(ab)^6 &= ab^5a^5b&&\\
\implies&& a(ba)^5b &=ab^5a^5b&&\\
\implies&& (ba)^5 &=b^5a^5&&\\
\implies&& (ba)^3(ba)^2 &=b^5a^5&&\\
\implies&& b^3a^3(ba)^2 &=b^5a^5&&[\text{by} (1)]\\
\implies&& a^3(ba)^2 &=b^2a^5&&\\
\implies&& a^3(ba)^2 &=a^5b^2&&[\text{by} (4)]\\
\implies&& (ba)^2 &=a^2b^2&&\\
\implies&& (ba)^2 &=b^2a^2&&[\text{by} (4)]\\
\implies&& baba &=b^2a^2&&\\
\implies&& ab &=ba&&\\[8pt]
\end{align*}
