Group and subgroup generated by element in G and application of Lagrange' s theorem

Question

It might be a stupid question.

Let G be a group and H be a subgroup

If there is "a" in H, then subgroup generated by a is a subgroup of H

Then, order of subgroup generated by a must divide the order of H

Since H is a subgroup, "a" is in G if "a" is in H.

So, subgroup generated by "a" is indeed a subgroup of G.

Then order of subgroup generated by a must divide the order of G

But, it seems to me that order of "a" must be common divisor of order of G and order of H.

Is it right? I am so confused!

Answer 1

For positive integers $k, l$ and $m$ it holds that if $k \mid l$ and $l \mid m$, then $k \mid m$ (divisibility is transitive). Now put $k=|\langle a \rangle|$, $l=|H|$ and $m=|G|$, then by Lagrange's Theorem (applied twice!) and the previous remark, $k \mid m$.