gcd and subgroup

Let H be a non trivial subgroup and a_1, a_2, \dots , a_n \in H. Prove that

    \[\gcd (a_1 , a_2 , \dots, a_n ) \mathbb{Z} \subseteq H\]

Solution

It is enough to prove it for the n=2 case since we can then reduce it by using the fact that

\gcd(a_1, a_2, \dots, a_n) = \gcd(a_1, a_2, \dots, a_{n-2}, \gcd(a_{n-1}, a_n))

We have that xa_1 + y a_2 \in H for any integers x and y. But by Bezout’s lemma we can find x , y such that

    \[x a_1 + y a_2 = \gcd(a_1, a_2)\]

and hence \gcd(a_1, a_2)\mathbb{Z} \subseteq H.

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