Is $\langle S \rangle \cup \langle T \rangle = \langle S \cup T \rangle$ and $\langle S \rangle \cap \langle T \rangle = \langle S \cap T \rangle$? Let $G$ a group, $S,T \subseteq G$ and $\langle S \rangle, \langle T ...--prophetes.ai

Is $\langle S \rangle \cup \langle T \rangle = \langle S \cup T \rangle$ and $\langle S \rangle \cap \langle T \rangle = \langle S \cap T \rangle$? Let $G$ a group, $S,T \subseteq G$ and $\langle S \rangle, \langle T \rangle$ the subgroups generated by $S$ and $T$. Is it true or false that $$\langle S \rangle \cup \langle T \rangle = \langle S \cup T \rangle$$ and $$\langle S \rangle \cap \langle T \rangle = \langle S \cap T \rangle?$$

Let $S=(12)$ and let $T = (13)$. Then $\langle (12)\rangle \cup\langle (13)\rangle$ has three elements (And is not even a group!).

But, $\langle (12), (13)\rangle = S_3$. This has 6 elements and IS a group.

You can do something quite similar for the intersection. The elements $S$ and $T$ should actually work as a counter example..

Explicitly: $\\{(12)\\}\cap\\{(13)\\} = \emptyset$ as a set intersection. But the intersection of $\langle (12)\rangle$ and $\langle (13)\rangle$ contains the identity.