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3. Let a, b be in G . If |a| and |b| are relatively prime (i.e. gcd (|a|, |b|)=1), then <a> intersection <b> ={e}. Prove the last statement.

Notice that $\displaystyle \left< a\right> \cap \left< b \right>$ is a subgroup of $\displaystyle \left< a \right>$ and $\displaystyle \left< b\right>$. By Lagrange's theorem it means $\displaystyle | \left< a\right> \cap \left< b\right>|$ divides $\displaystyle |\left< a\right>| = |a|$ and $\displaystyle \left< b\right>| = |b|$. Therefore, $\displaystyle |\left< a\right> \cap \left< b \right>| = 1 \implies \left< a \right> \cap \left< b \right> = \{ e \}$.