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$G$ 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.

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    As I recall you need to also assume that $3$ doesn't divide the order of $G$. See [this](http://math.stackexchange.com/questions/1179981/if-phig-g3-is-a-homomorphism-and-3-nmid-g-g-is-abelian) – 2017-02-14
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    But if $f$ is a ***monomorphism***, there cannot be elements of order $3$. Rather, should one assume $G$ to be finite? – 2017-02-14
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    I checked the original question. There is no such assumption. – 2017-02-14
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    @AndreasCaranti It does not have to be finite. For example $\mathbb Z$. – 2017-02-14
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    $a^3$ is not a homomorphism on $\mathbb{Z}$ – 2017-02-14
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    It has to be meant as $3 a$ in $\mathbb{Z}$. – 2017-02-14
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    oh right, of course. – 2017-02-14
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    @ziggurism: The group $\mathbb{Z}$ is a group under addition, so the correct interpretation of "$a^3$" is "$3a$". – 2017-02-14
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    @user150248, I was just wondering whether finiteness was necessary to prove the result in general. Indeed it is not, as I show in my answer below. – 2017-02-14
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    @user150248: based on your comment, it looks like my proof has a gap. Good catch! I'll delete it for now, and bring it back if I can fix it. Thanks for alerting me to the issue. – 2017-02-15
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    I fixed my previously flawed proof, so it's now undeleted. – 2017-02-18

2 Answers 2

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$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, $a b = ba$.

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    In hindsight this is actually not difficult – 2017-02-14
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    Hindsight is notoriously 20/20 ;-) Thanks. – 2017-02-14
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    I wish I knew how did you find the proof. – 2017-02-14
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    I have a long acquaintance with the subject. This comes with time, as you will see. – 2017-02-14
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    I shall try my best. Thank you. – 2017-02-14
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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*}

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    Why (ab)^6=a^6b^6? – 2017-02-15
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    Hmmm ... It looks like my proof has a gap. Good catch! I'll delete it for now (soon), and bring it back if I can fix it. Thanks for alerting me to the issue. – 2017-02-15
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    @user150248: All fixed!. If should be OK now. – 2017-02-18