What is an example where $G(p)/H(p) \not \cong (G/H)(p)$? Let $G$ be a finite abelian group and $H \subset G$ a subgroup. Let $G(p)$ be the $p$-torsion subgroup of $G$ for a prime $p$. Can we conclude that
$$
G(p)/H(p) \cong (G/H)(p)
$$
I've been told this is false, but the few examples I did by hand all work out. I'm probably just not trying anything big enough. Can someone provide a concrete example of where this fails? Or if it is true, can someone provide a proof?
I'm sure someone has already asked this, but I don't know the right words to search to find this. 
EDIT: The answers given to the previous question are excellent. However, they made me realize that I gave the wrong definition. I meant to define $G(p)$ as
$$
G(p) = \{g  \in G: \exists n, \; g^{p^n} = 1\}
$$
If we define $G(p)$ in this way, is the isomorphism still false?
 A: Take $G=\mathbb{Z}/4\mathbb{Z} = \{0,1,2,3\}$, take $H=\{0,2\}\cong \mathbb{Z}/2\mathbb{Z}$.
It is easy to see that $G(2)=H(2)=H$ and therefore $G(2)/H(2)$ is trivial.
However $G/H\cong \mathbb{Z}/2\mathbb{Z}$ and so $(G/H)(2)=G/H$.
Note that this case is for $p=2$ and can be generalized for every prime.
Edit: Answer to the edited question. Now the isomorphism holds.
I define an injection $\varphi:G(p)/H(p)\rightarrow (G/H)(p)$ as follows:
Let $gH(p)\in G(p)/H(p)$ and choose any representative $g\in G(p)$ I define $\varphi(g) = gH$.
$\varphi$ is well defined
First of all if $g\in H(p)$ then $g\in H$ and it sends to the trivial element. It is left to show that $gH\in G/H(p)$. By definition there exists $p^n$ such that $g^{p^n} = 1$ and therefore  $(gH)^{p^n} = g^{p^n}H=H$ is the unit element.
$\varphi$ is one to one
Let $gH(p),g'H(p)$ be such that $\varphi(g)=\varphi(g')$ then $gH=g'H$ and so $gg'^{-1}\in H$. Also as $g$ of order $p^n$ and $g'$ of order $p^{n'}$ we have that $gg'^{-1}$ is of order $n+n'$ and so belongs to $H(p)$ It follows that $gH_p=g'H_p$.
Therefore $G(p)/H(p)$ is isomorphic to a subgroup of $(G/H)(p)$. Now you can prove using cardinality arguments that this subgroup must be everything.
By the structure of finite abelian groups you can write
$$G\cong \bigoplus_{i=1}^m \mathbb{Z}/{p^{n_i}_i}\mathbb{Z}$$ for some fintie $m$ and possibly non-distinct $p_i$.
As $H$ is a subgroup it is isomorphic to
$$H\cong \bigoplus_{i=1}^m \mathbb{Z}/{p^{k_i}_i}\mathbb{Z}$$
for some $k_i\leq n_i$
It follows that $G/H\cong \bigoplus_{i=1}^m \mathbb{Z}/{p^{n_i-k_i}_i}\mathbb{Z}$
From this it is easy to see that the size of $G(p)/H(p)$ is equal to the size of $(G/H)(p)$ (which is equal to $\prod_{\{i:p_i=p\}} p^{n_i-k_i}$). Therefore they're isomorphic. In other words $\varphi$ is an isomorphism.
Edit 2: Answer to the question in the comment. The answer is yes (obviously) but it's not that nice...
We begin with $gH$ in $(G/H)(p)$ so as you said $g^{p^n}\in H$. We are going to modify $g$.
First by the structure of finite abelian groups we can write $H=\mathbb{Z}/p^m\mathbb{Z}\oplus T$ where $|T|$ is coprime to $p$. It is thus easy to see that the element $g^{p^{n+m}}\in T$.
Now the order of $T$ is coprime to $p$ we can find a $p^{n+m}$'th root for $g$, we call it $t\in H$ and it satisfies $g^{p^{n+m}}=t^{p^{n+m}}$. In particular $gt^{-1}\in G(p)$.
As $t\in H$ we have that $gt^{-1} H = gH$. We can therefore define the inverse of $\varphi$ to be the map $gH\mapsto gt^{-1}H(p)$. Note that this does not depend on the choice of $t$ for if $t'$ is also satisfying that $g^{p^{n+m}}=t'^{p^{n+m}}$ then $tt'^{-1}\in H(p)$. 
A: Let $p$ be a prime number; consider the finite abelian group $G=\mathbf Z/p^2\mathbf Z$ and its subgroup $H=p\mathbf Z/p^2\mathbf Z$. Then:


*

*$G(p)=H=H(p)$, so $G(p)/H(p)=\{0\}$;

*$G/H\simeq\mathbf Z/p\mathbf Z$, so $(G/H)(p)=G/H\ne\{0\}$.

