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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

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"order means the size of the group" - I don't know where you're comment just went, but okay. I thought that what the word cardinality was for though. – Ozera Feb 13 '14 at 16:26
up vote 2 down vote accepted

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.

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So $\left [ 1,2,3,5,6,10,15,30 \right ]$ are the orders of the subgroups of $Z_{30}$. A subgroup of order $m$ is generated by $a^{\frac{n}{m}}$. So for example we pick the subgroup of order $2$, then $a^{\frac{n}{m}} = a^{15}$. So..shouldn't this be $\left \{ 1,2,3,\cdots, a^{15}\right \}$? But that isn't of order $2$.. – Ozera Feb 13 '14 at 18:07
@Ozera: Since we're using additive notation, we'd rather write $a^{15}$ as $15a$, or just 15. The subgroup generated by $15$ is just the multiples of 15, i.e., $\{0,15\}$, and has order 2. – Kundor Feb 13 '14 at 18:21
@Kundor Ah, the multiples of $15$. Of course. Thank you, I understand now. – Ozera Feb 13 '14 at 18:29
Does this all make sense now, Ozera? You'll have a subgroup of order 3 (multiples of 10), of order $5$ (multiples of 6), etc. – amWhy Feb 13 '14 at 18:31
@Mike Thanks! I've edited accordingly. – amWhy Feb 13 '14 at 21:21


$$\begin{align*}&\;\;\left(\Bbb Z_{30}\,,\,+_{30}\right)\;\;\text{is a cyclic (additive) group}\\{}\\ &\;\;\text{In any cyclic group of order $\;n\;$, for any divisor $\;d\;$ of $\;n\;$ there exists a unique subgroup (also cyclic, of course) or order $\;d\;$ }\end{align*}$$

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Exactly! Thanks, this along with the other answers helped me complete the problem. – Ozera Feb 13 '14 at 21:13

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.

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