17
$\begingroup$

Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$

Then how prove $G$ is an abelian group?

Some context: Some of these commutation relations often imply that $G$ is abelian, for example if $(ab)^i = a^i b^i$ for three consecutive integers $i$ then $G$ is abelian, or if $g^2 = e$ for all $g$ then $G$ is abelian. This looks like another example of this phenomenon, but the same techniques do not apply.

$\endgroup$
  • 2
    $\begingroup$ from which text did you get this problem ? $\endgroup$ – Fawzy Hegab Mar 10 '13 at 19:32
  • $\begingroup$ also , is m,n are fixed or you mean , for all m,n in Z such that gcd(m,n)=1 .... ??? $\endgroup$ – Fawzy Hegab Mar 10 '13 at 19:39
  • $\begingroup$ @ MrWhymy:m,n are fixed integer number $\endgroup$ – M.H Mar 10 '13 at 19:46
  • $\begingroup$ it's nice problem , and the solution which is providen in the answer below is great :) $\endgroup$ – Fawzy Hegab Mar 10 '13 at 19:53
9
$\begingroup$

Let $M \subset G$ be the subgroup generated by all $m$-th powers and let $N \subset G$ be the subgroup generated by all $n$-th powers. These subgroups are clearly abelian normal subgroups. Since $m$ and $n$ are coprime, $G = MN$, and hence $M \cap N$ is contained in the center $Z(G)$ of $G$. To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] \subset M \cap N$ (since $M$ and $N$ are normal subgroups). Let $a \in M$ and $b \in N$. Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.

$\endgroup$
  • $\begingroup$ Dear Nicky! Let me ask sime questions: 1) Why $G=MN$? 2) Why $M\cap N$ is contained in the center $Z(G)$? $\endgroup$ – ZFR May 1 '18 at 14:21
  • 2
    $\begingroup$ (1) $gcd(m,n)=1$ whence you can find integers $k$ and $l$ with $1=km+ln$. So if $g \in G$ then $g=g^1=(g^k)^m \cdot (g^l)^n \in MN$. (2) Because $G=MN$ and $M$ and $N$ are abelian, any element from $M \cap N$ commutes with any element from $G$. Hope it is clear now. $\endgroup$ – Nicky Hekster May 1 '18 at 15:33
  • $\begingroup$ Dear Nicky! Thanks for answer! I got your solution and your approach is very simple and nice! +1 $\endgroup$ – ZFR May 1 '18 at 19:53
  • $\begingroup$ Thank you very much! $\endgroup$ – Nicky Hekster May 1 '18 at 20:32
3
$\begingroup$

Write $mk+nl=1$ for some $k,l\in\mathbb Z$.

We have $$ab=a^{mk+nl}b^{mk+nl}=a^{mk}(a^{nl}b^{mk})b^{nl}\stackrel{(*)}=a^{mk}(b^{mk}a^{nl})b^{nl}=(a^{mk}b^{mk})(a^{nl}b^{nl})=(b^{mk}a^{mk})(b^{nl}a^{nl})=b^{mk}(a^{mk}b^{nl})a^{nl}\stackrel{(**)}=b^{mk}(b^{nl}a^{mk})a^{nl}=(b^{mk}b^{nl})(a^{mk}a^{nl})=b^{mk+nl}a^{mk+nl}=ba,$$ where we used $$a^{nl}b^{mk}=b^{mk}a^{nl}\qquad(*)$$ and $$a^{mk}b^{nl}=b^{nl}a^{mk}\qquad(**)$$

Let's prove $(*)$ and $(**)$:

$$(a^mb^n)^{mk}=a^m(b^na^m)^{mk-1}b^n=a^m(b^na^m)^{mk}a^{-m}\Rightarrow(a^mb^n)^{mk}a^m=a^m(b^na^m)^{mk}\Rightarrow a^m(a^mb^n)^{mk}=a^m(b^na^m)^{mk},$$ hence $$(a^mb^n)^{mk}=(b^na^m)^{mk}.\qquad(\sharp)$$

Similarly we get $$(a^mb^n)^{nl}=(b^na^m)^{nl}.\qquad(\sharp\sharp)$$

Multiplying $(\sharp)$ and $(\sharp\sharp)$ we obtain $a^mb^n=b^na^m$.

Proceeding in the same way we get $a^nb^m=b^ma^n$

An easy induction shows now $(*)$ and $(**)$.

$\endgroup$
-5
$\begingroup$

As $\gcd(m, n) = 1$, by Bezóut's identity there are $u$, $v$ such that $u m + v n = 1$.

Take $a, b \in G$ arbitrary. By the conditions, as $a^u \in G$ for all $u \in \mathbb{Z}$, and as the conditions allow to conmute certain powers: $$ \begin{align*} a^{u m} b^{u m} &= b^{u m} a^{u m} \\ (a^{u m} b^{u m})^{v n} &= a^{u m + v n} b^{u m + v n} \\ &= a b \\ (b^{u m} a^{u m})^{v n} &= b^{u m + v n} a^{u m + v n} \\ &= b a \end{align*} $$ This because e.g. $(a^{u m} b^{u m})^2 = a^{u m} b^{u m} a^{u m} b^{u m} = a^{u m} a^{u m} b^{u m} b^{u m} = a^{2 u m} b^{2 u m}$ since the middle pair conmutes by the conditions. Now use induction.

(Perhaps the switching around can be expressed in a more compact way).

Nice problem!

$\endgroup$
  • 1
    $\begingroup$ Great Solution :) thanx :) $\endgroup$ – Fawzy Hegab Mar 10 '13 at 19:52
  • 4
    $\begingroup$ @vonbrand:why we have $(b^{u m} a^{u m})^{v n} = b^{u m + v n} a^{u m + v n}$? $\endgroup$ – M.H Mar 10 '13 at 20:30
  • $\begingroup$ @MaisamHedyelloo , we have a in G , so $a^u$ is in G , choose u to be the same u in the relation $mu+vn=1$ , apply the hyposetes to $a_1$ = $a^u$ and use induction as it's mentioned in the answer in the speicial case for 2 . $\endgroup$ – Fawzy Hegab Mar 10 '13 at 20:45
  • 2
    $\begingroup$ How do you get $(a^{u m} b^{u m})^{v n} = a^{u m + v n} b^{u m + v n}$? Doesn't $(a^{u m} b^{u m})^{v n} = a^{u m v n} b^{u m v n}$? $\endgroup$ – Code-Guru Mar 10 '13 at 22:11
  • $\begingroup$ @MrWhy Huh? What's $a_1$? $\endgroup$ – Code-Guru Mar 10 '13 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.