Let \(G\) be an abelian group. Suppose \(a,b\in G\) have finite orders \(r,s\) with \[ \gcd(r,s)=1. \] Show that \(ab\) has order \(rs\).
Solution
1. The element \(ab\) satisfies\[ (ab)^{rs}=e. \]
Since \(G\) is abelian, \[ (ab)^{rs} = a^{rs}b^{rs} = (a^r)^s(b^s)^r =e. \] Thus \(ab\) has finite order dividing \(rs\).
2. If\[ (ab)^m=e \]for some positive integer \(m\), then\[ r\mid m. \]
Commutativity gives \[ a^mb^m=e, \] so \[ a^m=b^{-m}. \] Raise both sides to the \(s\)th power: \[ a^{ms} = b^{-ms} = (b^s)^{-m} =e. \] Since \(a\) has order \(r\), this implies \(r\mid ms\). Because \(\gcd(r,s)=1\), Euclid’s lemma gives \(r\mid m\).
3. Under the same hypothesis,\[ s\mid m. \]
From \(a^m=b^{-m}\), raise both sides to the \(r\)th power: \[ (a^r)^m=b^{-mr}. \] Hence \(b^{mr}=e\). Since \(b\) has order \(s\), one has \(s\mid mr\). The coprimality \(\gcd(r,s)=1\) therefore gives \(s\mid m\).
4. Every positive exponent \(m\) satisfying \((ab)^m=e\) is divisible by\(rs\).
Steps 2 and 3 give \[ r\mid m \qquad\text{and}\qquad s\mid m. \] Since \(\gcd(r,s)=1\), their product divides \(m\): \[ rs\mid m. \]
5. Therefore\[ \boxed{\operatorname{ord}(ab)=rs}. \]
Step 1 shows that the order of \(ab\) divides \(rs\). Step 4 shows that every positive exponent annihilating \(ab\), in particular its order, is divisible by \(rs\). The two divisibilities force equality.
6. Q.E.D.
Step 5 is the required conclusion.