Find the number of the subgroups Here is the question that I've stuck.
Question)
$G = Z_{100} \times Z_{500}$
Let $H$ is a subgroup of the $G$ 
How many number of the $H$   s.t. $H \simeq Z_{25} \times Z_{25}$ ?
My idea) This is my idea when I tried this question at the first time.
The ideal is a really simple. 
Let me suppose only finite group case.
Put the $g (\in G)$ $s.t.$ $\vert g \vert = m$.
Since there are some elements that generate the same subgroups
Then All we have to do is 
(The number of the $H$ whose order is $m$)  = 
(The number of the $g$) / (The number of the group's element by generated by element, $g$)
But, I couldn't find any the denominator in formula of the above question.
 A: Here's one way to prove that there is just one subgroup without thinking too much about it. You have an abelian group $G$ (the particular primes involved in the factorisation of $|G|$ don't matter) and you want to find how many $p$-subgroups of some type it contains. Since $G$ has a unique Sylow $p$-subgroup $P$, every such $p$-subgroup is contained in $P$ thus the problem translates into one about counting $p$-subgroups of given type inside an abelian $p$-group.
I wrote all that just to say that it suffices to count the number of subgroups of $P = C_{25} \times C_{125}$ of type $C_{25} \times C_{25}$. Ignoring the particular prime $p=5$, we want to count the number of subgroups of $P = C_{p^2} \times C_{p^3}$ of type $C_{p^2} \times C_{p^2}$.
There is a standard theorem which allows you to do that. Call $H$ of type $\mu = (\mu_1,\ldots,\mu_k)$ if $H= C_{p^{\mu_1}} \times \ldots \times C_{p^{\mu_k}}$. The theorem says that the number of subgroups of type $\nu = (\nu_1,\ldots,\nu_\ell)$ 
in an abelian $p$-group of type $\lambda = (\lambda_1, \ldots, \lambda_\ell)$ is given by the formula
$$\prod_{i\geq 1} p^{\nu_{i+1}'(\lambda_i' - \nu_i')} {\lambda_i' - \nu_{i+1}' \brack \nu_i' - \nu_{i+1}'}_p,
$$
where $\lambda', \nu'$ are the conjugates of the partitions $\lambda$ and $\nu$, respectively, and
\begin{equation}
{n \brack k}_p = \prod_{i=0}^{k-1} \frac{1 - p^{n-i}}{1-p^{i+1}}
\end{equation}
is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the field $\mathbb{Z}/p\mathbb{Z}$; 
see, for instance, this (Eqn. ($1$)). 
In our example, we have $\nu =(2,3)$, $\nu' = (2,2,1)$ and $\lambda=\lambda'=(2,2)$. If you do the algebra you'll see that the answer is $1$.
This whole thing is very likely overkill, but at least it allows you to solve this and  other similar problems uniformly.
A: $\Bbb Z_{100}×\Bbb Z_{500}\cong(\Bbb Z_{25}×\Bbb Z_{125})×(\Bbb Z_4×\Bbb Z_4)$.  
Since when the orders of the factors  are relatively prime the subgroup of a product is a product of subgroups (see this),  we can consider subgroups of $\Bbb Z_{25}×\Bbb Z_{125}$ of order $625$.
The only possibilities are:   $\Bbb Z_{25}×\Bbb Z_{25}$ and $\Bbb Z_5×\Bbb Z_{125}$.
So only one.
