Groups and Subgroups - Confusion Find all subgroups (and their orders) of the group $<\mathbb{Z}_{30}, +_{30}>$.
This material is new to me. I know what a subgroup is by reading the definition, but what does the problem mean by 'orders'? So, for the definition of a subgroup, if H $\subseteq G$ (and $(G, *)$ is a group) and $H$ is a group with the same binary operation $*$, then $H$ is called a $\textit{subgroup}$.
But.. how do I go about finding every single subgroup? And what is meant by 'orders'?
Order == Cardinality == Size of the Set/Group
Right, now how do we find all of the subgroups? 
RESOLVED - See comments
 A: Hints:
$$\begin{align*}&\;\;\left(\Bbb Z_{30}\,,\,+_{30}\right)\;\;\text{is a cyclic (additive) group}\\{}\\
&\;\;\text{In any cyclic group of order $\;n\;$, for any divisor $\;d\;$ of $\;n\;$ there exists a unique subgroup (also cyclic, of course) or order $\;d\;$ }\end{align*}$$
A: By the Fundamental Theorem of Finitely Generated Abelian Groups The group $\mathbb{Z}_{30}$ is isomorphic to
$$
\mathbb{Z}_2\times \mathbb{Z}_3\times \mathbb{Z}_5
$$
since $30=2\cdot 3\cdot 5$.  You can find all the subgroups by taking factors.  The "order" of a group is just how many elements are in it. 
A: Any group of the form $\langle \mathbb Z_n, +_n\rangle$ is cyclic (meaning there is some $g\in \mathbb Z_n$ such that $g$ generates the entire group: $\langle g\rangle = \mathbb Z_n)$.
The order of $\mathbb Z_n$ is $n$, meaning it contains exactly $n$ elements.
A subgroup with $m$ elements has order $m$. 
Every subgroup of a cyclic group is also cyclic, meaning that if $H$ is a subgroup of $\mathbb Z_n$, then there is an element $h\in H$ such that $h$ generates all of $H$.
For a cyclic group of order $n$, there exists a unique (one and only one) subgroup of order $m$ for every $m$ that divides $n$, and for any $r$ such that $r$ does NOT divide $n$, no subgroup of order $r$ exists. 
So in the case of $\mathbb Z_{30}$, since $1, 2, 3, 5, 6, 10, 15, 30$ all divide $30$, you can expect to find for each divisor $m$, a unique subgroup whose order is equal to $m$.
Of course, $\{0\}$ is the trivial subgroup of order $1$, and $\mathbb Z_{30}$ is a subgroup of itself, and of order $30$. 
Six additional subgroups of $\mathbb Z_{30}$ exist. 
