Showing that intersection of cyclic groups is identity

Question

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!

The intersection of subgroups is a subgroup. And according to Lagrange theorem the order of a subgroup divides the order of the group. — mathcounterexamples.net, Nov 02 '15 at 22:32
The order of the intersection divides 10 and 21, so it divides the greatest common divisor, that is 1. Hence the intersection has order 1, thus it's the identity. — user279325, Nov 02 '15 at 22:32
Probably here $|a|$ means the order of $a$. In general one uses the Bezout Identity, but here we can note that $21+(-2)(10)=1$. — André Nicolas, Nov 02 '15 at 22:33
{3} is not a group and |3| $\ne$ 1. The only group of order 1 is {e}. The intersection of subgroups is a subgroup. The intersection of cyclic subgroups is a cyclic subgroup. If c is in both and then |c| divides 10 and |c| divides 21 so |c| = 1. Thus c is such that $c^1 = e$. $3^1 = 3 \ne e$. So $c \ne 3$. In fact $c^1 = c$ so $c = c^1 = e$ is the only member of the group. — fleablood, Nov 02 '15 at 22:59
Because ${3}$ is not a subgroup. Note that identity doesn't belong to it. — user279325, Nov 02 '15 at 23:00
The "order" of an element is different than the "order" of a set. The order of a set or group is the number of elements. The order of a group element, a, is the smallest power, m, at which $a^m = e$. The order of a group element always divides the order of the group set. {3} is a set but it is not a group. To make a group that has 3 in it, you need $<3> = {e, 3, 3^2, ....}$. The order of 3 and the order of <3> is not 1. — fleablood, Nov 02 '15 at 23:05

score 4 · Accepted Answer · answered Nov 03 '15 at 04:16

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 $.

$<a> \cap $ is a group. Let $c \in <a> \cap $. Then $c \in <a>$ and $c \in $ 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 $.

Showing that intersection of cyclic groups is identity

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