If a group has commutative generators is the group always abelian?

I have a question dealing with how to determine if a Cayley graph of a group is an abelian group. It seems that if the generators commute then the entire group will be abelian because all elements will be a "power" of the generators.

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    $\begingroup$ Yes. The proof is by induction. To commute two elements of the group, repeatedly commute the generators that make them up. $\endgroup$ Jan 25, 2015 at 21:11
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    $\begingroup$ This is not really a soft question. I've edited the tags accordingly. $\endgroup$ Jan 25, 2015 at 21:11
  • $\begingroup$ @MattSamuel okay thanks, I guess I wasn't really sure what deemed a question soft or not. $\endgroup$ Jan 25, 2015 at 21:15

1 Answer 1


Let $G$ be our group with generating set $(a_i)_{i\in I}$ where $I$ is some index set and suppose all $a_i$ commute with each other. If $x\in G$ is an element, then there are indices $i_1,i_2,\ldots,i_m$ such that $x=a_{i_1}a_{i_2}\cdots a_{i_m}$. Let $x,y\in G$ and suppose that $x=a_{i_1}a_{i_2}\cdots a_{i_m}$ and $y=a_{j_1}a_{j_2}\cdots a_{j_n}$. We prove that $xy=yx$ by induction on $n$. If $n=0$ this means that $y$ is the identity, which commutes with every element.

Next we prove the result for $n=1$. This we do by induction on $m$. For $m=0$ $x$ is the identity, so the result holds. For $m>0$ we have $$xy=a_{i_1}a_{i_2}\cdots a_{i_m}a_{j_1}=a_{i_1}a_{i_2}\cdots a_{i_{m-1}}a_{j_1}a_{i_m}$$ due to the fact that $a_{j_1}$ and $a_{i_m}$ commute, and since by the induction hypothesis $a_{j_1}$ commutes with $a_{i_1}a_{i_2}\cdots a_{i_{m-1}}$ it follows that $$xy=(a_{i_1}a_{i_2}\cdots a_{i_{m-1}})a_{j_1}a_{i_m}=a_{j_1}(a_{i_1}a_{i_2}\cdots a_{i_{m-1}})a_{i_m}=yx$$ so the result follows for $n=1$.

Assume now that $n>1$. Then we have $$xy=(a_{i_1}\cdots a_{i_m})a_{j_1}(a_{j_2}\cdots a_{j_n})=a_{j_1}(a_{i_1}\cdots a_{i_m})(a_{j_2}\cdots a_{j_m})=a_{j_1}(a_{j_2}\cdots a_{j_n})(a_{i_1}\cdots a_{i_m})=yx$$ The first equality holds since we have proved the result for $n=1$ and the second equality holds by the induction hypothesis. Thus it follows by induction that $xy=yx$ for all $x,y\in G$, and therefore $G$ is abelian.


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