What are the subgroups of $Z_{3^2}$ Need to find all the subgroups of $Z_{3^2}$. Is $Z_{3^2}$ same as the Elementary abelian group (the product $Z_3\times Z_3$)?
 A: $\mathbb Z_{3^2} = \mathbb Z_{9} \neq \mathbb Z_3\times \mathbb Z_3$ and clearly, $\mathbb Z_9$ cyclic (and so all its subgroups are cyclic). 
Now, use the fact that for cyclic groups of order $n$, the converse of Lagrange's Theorem holds: For every positive integer $m$ such that $m\mid n$, there exists a unique subgroup of order $m$. 
So to find the subgroups of $\mathbb Z_9$, for each positive integer that divides 9, determine the corresponding subgroup: there are exactly three integers dividing $9$: $1, 3, 9$. The subgroup of order $1$ is of course, the trivial group $\{0\}$. The subgroup of order $3$ must be $\mathbb Z_3$, and the subgroup of order $9$ is $\mathbb Z_9$ itself.
A: For any $n$, the set of all subgroups of $\mathbb{Z}_n = \langle x \rangle$ can be shown to be the set $\{ \langle x^d \rangle: d | n\}$.  Every subgroup of a cyclic group is a cyclic group generated by the smallest positive power (denoted by $x^d$ here) of $x$ that is in the subgroup.  Notice that $\langle x^d \rangle$ is isomorphic to $\mathbb{Z}_{n/d}$, and that as $d$ runs over the divisors of $n$, so does $n/d$.  Thus, the set of all subgroups of $\mathbb{Z}_n$ is isomorphic to  $\{\mathbb{Z}_d: d | n\}$.  More specifically, the subgroups of $\mathbb{Z}_9=\langle x \rangle$ are $\langle x^d \rangle$, where $d$ equals 1, 3 or 9, which are the subgroups $\mathbb{Z}_9, \{1,x^3,x^6\} \cong \mathbb{Z}_3$, and the trivial group $\{1\}$, respectively. 
A: The lattice of subgroups of $\mathbb Z_m$ is isomorphic to the lattice of divisors of $m$ via the canonical homomorphism $\mathbb Z \to \mathbb Z_m$ whose kernel is $m\mathbb Z$.
Recall that the lattice of subgroups of $\mathbb Z$ is isomorphic to the lattice $\mathbb N$ under divisibility.
