group theory, cardinality of a quotient group Let $H$ be any finite abelian group. Define $H^n=\lbrace x^n \text{ }|\text{ }x\in H\rbrace$. It is easy to see that $H^n$ is a subgroup of $H$. It is also easy to see that $H^n\subseteq H^d$ if $d\text{ }|\text{ }n$. 
Is it true that all subgroups containing $H^n$ are of the form $H^d$ for some divisor of $n$ ?
Also, can we say that $|H/H^n|\leq n$ ?
 A: If $\;H=\Bbb Z_2\times\Bbb Z_2\;$ , then $\;H^2=\{(0,0)\}\;$  and the subgroup $\;K=\{\,(0,0),\,(1,1)\,\}\le H\;$ is not of the form $\;H^d\;$ , for $\;d=1,2\;$.
For second part: take $\;H=\Bbb Z_2\times\Bbb Z_4\;$ and $\;H^4=\{(0,0)\}\;$ , but
$$\left|H/H^n\right|=|H|=8>4$$
A: In the case $ H$  finite cyclic with $n\leq \mid H\mid $  the
first statement is ok, because $H=\langle x \rangle$ and
$H^n=\langle x^n \rangle\subseteq K=\langle x^d\rangle$ applique
that $n=md$ so $d\mid n$. but in the case abelian not cyclic, the
statement is false (the  counter examples as given above in
comment and in the answer).here  deduced a counter example,  by
examining what happens in the factorized
 abelian group as cyclic subgroup product: $H=Z/p^2Z\times Z/p^2Z$
 and $n=p$ then $H^p=Z/pZ\times Z/pZ$
if we tack $K=Z/p^2Z\times Z/pZ$ then we have $H^p\subseteq K$ but
$K$ can not wrote as $H^d$ (note: for eache composante of $K$
there are one d in this example $d_1=p$ for the first composant
and $d_2=1$ for the second composante).
 The second statement is qlso true in the  cyclic case because for any $n$
 the order of $x^n$ is $gcd(n,order(x))$ and so $ \mid H/H^n\mid =gcd(n,order(x))\leq n$.
 Also by examining the situation in the case abelian not cyclic  we see   for any $n$ that
 $ \mid H/H^n\mid $ is increased by $n^s$ where $s$ is the number of cyclic factors in the
 factorization
  of $H$ as product of cyclic  sub group.
as a counter example just tack
$H=Z/pZ\times Z/pZ$ if $n=p$ then  $ \mid H/H^p\mid=p^2
>p=n$.
