Showing that intersection of cyclic groups is identity Let $|a|=10, |b|=21$. Show that $\langle a\rangle\cap\langle b\rangle=\{e\}$.
I think that $|a|=10$ implies there are subgroups of the cyclic group of order $2,5$; $|b|=21$ implies that there are subgroups of order $3,7$. How can I proceed to conclude that their intersection is identity?
Thanks in advance!
 A: 1) If A, B are subgroups of G then $A \cap B$ is a group with the group operator.
Proof: If a, b $\in A \cap B$ Then $a,b \in A$ and $a,b \in B$ then $ab \in A $and $ab \in B$ (because A and B are groups) so $ab \in A \cap B$.  So the group operator is closed in $A \cap B$. 
Likewise $e \in A$ and $e \in B$ so $e \in A \cap B$.  And for all $a \in A \cap B$ $a^{-1}$ is in both A and B so $a^{-1} \in A \cap B$ so $A \cap B$ is a group.
2)  If $a$ is an element of a group with |a| = m (ie $a^m = e$) then the order of the element $b = a^k$ must divide m.
Proof: $b^m = (a^k)^m = (a^m)^k = e^k = e$.  Suppose |$b$| = $v$. And suppose $m = vj + i; 0<= i <v$.  Then $e = b^m = b^{vj + i} = (b^v)^jb^i = eb^i = b^i.$ But  $|b| = v > i$ so $i$ must be equal to $0$. So m = vj, v = |b| divides m.
3) Let |a| = 10, |b| = 21, the $\{e\} = <a> \cap <b>$.
$<a> \cap <b>$ is a group. Let $c \in <a> \cap <b>$. Then $c \in <a>$ and $c \in <b>$ so |c| divides |a| = 10 and |c| divides |b| = 21.  So |c| divides gcd(10, 21) = 1.  So |c| = 1.  So $c^1 = c = e$.  Thus $e$ is the only element of $<a> \cap <b>$.
