Group theory question. If $m$ and $n$ are relatively prime, and $a$, $b$ belong to group and $a^m = b^m$ and $a^n = b^n$, how would one prove $a = b$?

Question

How would you prove that 2 elements of the group $G$ are equal, i.e., $a = b$ if we know that $a$, $b$ belong to group $G$ and that $m$, $n$ are relatively prime (integers) and that $a^m = b^m$ and $a^n = b^n$

Thanks!

Answer 1

Since $m$ and $n$ are relatively prime, by Bezout's lemma there exist $x$, $y \in \mathbb{Z}$ such that $xm + yn = 1$. Then, it follows $$a = a^{xm+yn} = (a^m)^x (a^n)^y = (b^m)^x (b^n)^y = b^{xm+yn} = b.$$

Answer 2

From Bezout's theorem, there exist integers $h$ and $k$ such that $hm + kn = 1$. Then $$(a^m)^h(a^n)^k = (b^m)^h(b^n)^k \implies a^{hm + kn} = b^{hm + kn} \implies a = b$$

Answer 3

Consider the subgroup $H=\{\,k\mid a^k=b^k\,\}$ of $(\Bbb Z,+)$ (it clearly contains $0$ and is closed under subtraction: if $a^k=b^k$ and $a^l=b^l$, then $a^{k-l}=a^k(a^l)^{-1}=b^k(b^l)^{-1}=a^{k-l}$). By hypothesis $H$ has $m,n$ as elements, an these being relatively prime means that $H$ must be all of $\Bbb Z$ (standard result that subgroups of $\Bbb Z$ are cyclic; a generator must divide $m$ and $n$). Hence $1\in H$, which proves the result.