Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

List all the elements of the subgroup generated by the subset $\{2,3 \}$ of $\mathbb{Z}_{12}$

The solution said $\langle \gcd(2,3,12)\rangle = \langle 1\rangle = \mathbb{Z}_{12}$

Could someone explain to me why are we taking the gcd?

What exactly does it mean for $\{2,3 \}$ to generate things? I am not seeing any connection or picture here

share|cite|improve this question
If you write \gcd rather than gcd (with no backslash) then not only will the letters not get italicized like variables, but proper spacing will appear in expressions like $a\gcd(b,c)$. Also, I change $<1>$ to $\langle 1 \rangle$ (coded as \langle 1 \rangle). That's also standard usage. – Michael Hardy Nov 22 '12 at 3:58
up vote 1 down vote accepted

If $G$ is a group and $S$ is a subset of $G$, $\langle S\rangle$, the subgroup generated by $S$, is simply the smallest subgroup of $G$ that contains every element of $S$. It can be defined as $\bigcap\{H\leqslant G:S\subseteq H\}$, the intersection of all subgroups of $G$ containing $S$ as a subset. However, it’s probably more instructive to look at it a bit differently.

Consider the subset $S=\{4,6\}$ of $\Bbb Z_{12}$. $S$ is not a subgroup of $\Bbb Z_{12}$: it’s not closed under addition, it doesn’t contain the identity element, and it doesn’t contain the additive inverse of $4$. What, at a bare minimum, do we have to add to $S$ to turn it into a subgroup of $\Bbb Z_{12}$? We certainly have to throw in $-4=8$, though $-6=6$ is already in $S$; that expands $S$ to $\{4,6,8\}$. We also have to close the set under addition, so we have to throw in $4+6=10$, $4+8=0$, and $6+8=2$, bringing the set up to $\{0,2,4,6,8,10\}$. This is a subgroup of $\Bbb Z_{12}$, so nothing more need be added: $$\big\langle\{4,6\}\big\rangle=\{0,2,4,6,8,10\}\;.$$

$S=\{4,6\}$ generates the subgroup $\{0,2,4,6,8,10\}$ in the sense that we can work out from $S$ as I did above, adding elements as we discover that they’re needed if we’re to have a group, until we’ve added just enough to get a group, and the process ends.

Finding $\langle S\rangle$ when $S$ is a subset of one of the cyclic groups $Z_n$ turns out to be especially easy. That’s where the greatest common divisor comes in. Notice that in my example the subgroup of $\Bbb Z_{12}$ generated by $\{4,6\}$ turned out to be the set of multiples of $2$ in $\Bbb Z_{12}$, which is easily seen to be the subgroup of $\Bbb Z_{12}$ generated by the single element $2$: once you have $2$, you must have $2+2=4$, $4+2=6$, $6+2=8$, $8+2=10$, and $10+2=0$, and these six elements do indeed form a subgroup of $\Bbb Z_{12}$. It’s not an accident that $2=\gcd\{4,6,12\}$; in general is you have a subset $S\subseteq Z_n$, you’ll find that $\langle S\rangle$ is just the set of multiples (in $\Bbb Z_n$) of the gcd of $n$ and the members of $S$. This is a matter of elementary number theory; I suspect that it’s been proved either in class or in your text.

share|cite|improve this answer

The subgroup generated by a set $A$ is the smallest subgroup that contains all elements in $A$. So you just put all possible combinations of elements in $A$. But for abelian groups like $\mathbb{Z}_{12}$ it is easier, the subgroup generated by $\{2,3\}$ is the collection $\{n\cdot 2+m\cdot 3:n,m\in\mathbb{Z}\}$.

This is also when gcd comes into play. Because $\operatorname{gcd}(2,3)=1$, we know there are $x,y\in \mathbb{Z}$ such that \begin{equation} x\cdot 2+y\cdot 3=1, \end{equation}so $1$ is in the subgroup generated by $\{2,3\}$. But $1$ is one generator of $\mathbb{Z}_{12}$, so once you have $1$, you have everything. That is why the subgroup is actually the entire $\mathbb{Z}_{12}$.

share|cite|improve this answer

It means to take all the integer multiples of $\{2,3\}\,$ and their sums, since the operation in $\,\Bbb Z_{12}\,$ is additive. If it were multiplicative it'd mean to take all the powers and products of those elements.

In this case:

$$(-1)\cdot 2+1\cdot 3=1\Longrightarrow \forall\,k\in\Bbb Z_{12}\,\,,\,k=k\cdot 1=(-k)\cdot 2+k\cdot 3$$

so $\,k\,$ belongs to the subgroup generated by $\,\{2,3\}\,$ and thus this subgroup is, in fact, the whole group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.