# Thread: Union of directed family of subgroups is a group

1. ## Union of directed family of subgroups is a group

Show that the union of a nonempty directed family of subgroup of a group G is a subgroup of G.

Proof so far:

Let G be a group, and suppose that $\displaystyle \{ S _{a} \} _{a \in A}$ be a directed family of subgroups of G.

Then $\displaystyle \bigcup \{ S _{a} \} _{a \in A}$ is the union.

Since it is nonempty, the subgroup of {e}, act as the identity of this subgroup.

Now, suppose that x and y are elements of this union, then there exist i,j in A such that $\displaystyle x \in S_{i}$ and $\displaystyle y \in S_{j}$. And since this family is directed, there exist some k in A such that $\displaystyle S_{i} \subseteq S_{k}$ and $\displaystyle S_{j} \subseteq S_{k}$. Implies that both x,y are in $\displaystyle S_{k}$ And since the union is consist of all subgroups, that means $\displaystyle S_{k}$ is a subgroup, implies that $\displaystyle xy^{-1} \in S_{k}$, thus proves the claim.

Is this right? Thank you.

2. Let $\displaystyle S = \cup_{i\in I} S_i$.
Here is how to start off.
If $\displaystyle x,y\in S$ then $\displaystyle x\in S_a$ and $\displaystyle y\in S_b$.
Therefore there is $\displaystyle c$ such that $\displaystyle a\leq c,b\leq c$.
And thus, $\displaystyle S_a \subseteq S_c$ and $\displaystyle S_b\subseteq S_c$.
Thus, $\displaystyle x,y\in S_c$ thus $\displaystyle xy\in S_c\implies xy\in S$.
Thus $\displaystyle S$ is closed.

Continue with the other properties.