# $\langle s \rangle$ As An Intersection Of Subgroups

I am acquainted with the following definition

Let $G$ be a group, $\emptyset\neq A\subset G$

$\langle A \rangle=\{x_1\cdot x_2 \cdots x_n: n\in \mathbb{N}\cup\{0\}\forall_{i} x_i\in A\cup A^{-1} \}$ where $A^{-1}$ is the family of all invertible elements of $A$

So when I am given $G=\mathbb{Z}$ and $\langle A \rangle=\{6,8\}$ I can quickly understand why $\langle A \rangle=\{6,8\}=2\mathbb{Z}$

But why looking at the definition $\langle A \rangle=\cap H_{H\leq G,A\subseteq H}$ How can I conclude that $\langle A \rangle=\{6,8\}=2\mathbb{Z}?$

• It actually defines $\langle A\rangle$, the subgroup generated by $A$, as the smallest subgroup of $G$, that contains the entire set $A$. Of course if $A\subseteq H$, for any subgroup $H$, then $\langle A\rangle\subseteq H$, because $H$ should satisfy all the properties for being a group. And obviously $A\subseteq \langle A\rangle$. Hence the definition $\langle A \rangle=\cap H_{H\leq G,A\subseteq H}$ – Abishanka Saha Dec 12 '17 at 16:27
• In a sense it is similar to the definition of a Span? Meaning it is the intersection of all subgroups of $G$ that are containing the elements of $\langle A \rangle$? – gbox Dec 12 '17 at 16:55
• Yeah, the same intuition works here too. – Abishanka Saha Dec 12 '17 at 17:01

Note that $8-6=2$, so if $\{6,8\}\subseteq H$ for any subgroup $H$ of $\Bbb{Z}$, then $2\in H$, and hence $2\Bbb{Z}\subseteq H$. And obviously $\{6,8\}\in2\Bbb{Z}$. So by the definition $\langle\{6,8\}\rangle=2\Bbb{Z}$