# Elementary subgroup question

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

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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

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.

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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 $$x\cdot 2+y\cdot 3=1,$$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 $\mathbb{Z}_{12}$.

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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.

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