# Show that union of ascending chain of subgroups is subgroup

• Mr Davis 97
In summary, the given conversation discusses the proof that the union of an ascending chain of subgroups in a group is also a subgroup of that group. The proof involves showing that the union is nonempty and that it satisfies the subgroup criteria.

## Homework Statement

Let ##H_1 \le H_2 \le \cdots## be an ascending chain of subgroups of ##G##. Prove that ##H = \bigcup\limits_{i=1}^{\infty} H_{i}## is a subgroup of ##G##.

## The Attempt at a Solution

Certainly ##H## is nonempty, since each subgroup ##H_i## has at least the identity element. Now, let ##a,b \in H##. Then ##a \in H_i## where ##i## is taken to be minimal. Also ##b \in H_j##, where ##j## is taken to be minimal. WLOG suppose that ##i \le j##. Then ##H_i \subseteq H_j## and so ##a,b \in H_j##. Then since ##H_j## is a subgroup, ##ab^{-1} \in H_j \subseteq H##, and so ##ab^{-1} \in H##.

Delta2
Mr Davis 97 said:

## Homework Statement

Let ##H_1 \le H_2 \le \cdots## be an ascending chain of subgroups of ##G##. Prove that ##H = \bigcup\limits_{i=1}^{\infty} H_{i}## is a subgroup of ##G##.

## The Attempt at a Solution

Certainly ##H## is nonempty, since each subgroup ##H_i## has at least the identity element. Now, let ##a,b \in H##. Then ##a \in H_i## where ##i## is taken to be minimal. Also ##b \in H_j##, where ##j## is taken to be minimal. WLOG suppose that ##i \le j##. Then ##H_i \subseteq H_j## and so ##a,b \in H_j##. Then since ##H_j## is a subgroup, ##ab^{-1} \in H_j \subseteq H##, and so ##ab^{-1} \in H##.
Correct.

Mr Davis 97