Answer to Question #350796 in Abstract Algebra for liz

Question

2.10. Let H be the subgroup generated by two elements a, b of a group G. Prove that if ab = ba, then H is an abelian group.

Answer 1

The elemtnts of $H$ are of the form: $a^{i_1}b^{i_2}a^{i_3}\cdots a^{i_k-1}b^{i_k}$ where $i_1,\dots, i_k\in\mathbb{Z}$ for some $k$.

So let $x,y\in H$. Then we can write $x=a^{i_1}b^{i_2}a^{i_3}\cdots a^{i_k-1}b^{i_k}$ and $y=a^{j_1}b^{j_2}a^{j_3}\cdots a^{j_k-1}b^{j_k}$. Then $xy=(a^{i_1}\cdots b^{i_k})(a^{j_1}\cdots b^{j_k})$ since $ab=ba$ we can interchange each term in this multiplication, and obtain: $xy=(a^{j_1}\cdots b^{j_k})(a^{i_1}\cdots b^{i_k})=yx$.

This implies $H$ is abelian.