# Conditions on H and K if H ∪ K is a subgroup

• Mr Davis 97
In summary, to prove that if ##H \cup K## is a subgroup of ##G## then ##H \subseteq K## or ##K \subseteq H##, we can assume that ##H \cup K \le G## and then show that if neither H nor K is a subset of the other, it leads to a contradiction. This is because if ##H \cup K## is a subgroup, then ##hk \in H \cup K## for all ##h \in H## and ##k \in K##. Therefore, we can conclude that either ##h \in H## and ##k \in K##, or ##k \in H## and ##h \in K##, contradicting our
Mr Davis 97

## Homework Statement

Let H and K be subgroups of G. Prove that if ##H \cup K## is a subgroup of ##G## then ##H \subseteq K## or ##K \subseteq H##

## The Attempt at a Solution

Suppose that ##H \cup K \le G##. For contradiction, suppose that neither H nor K is a subset of the other. This implies that there exists an ##h \in H## that is not in ##K## and a ##k \in K## that is not in ##H##. Since ##H## and ##K## are individually subgroups, ##h^{-1} \in H## and ##k^{-1} \in K##. Now since ##h \in H \cup K## and ##k \in H \cup K##, we know that ##hk \in H \cup K##, since we supposed it's a subgroup. We have three cases to check: If ##hk \in (H \cap K)##, then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - H## then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - K## then ##h^{-1}(hk) = k \in H##, a contradiction.

Mr Davis 97 said:

## Homework Statement

Let H and K be subgroups of G. Prove that if ##H \cup K## is a subgroup of ##G## then ##H \subseteq K## or ##K \subseteq H##

## The Attempt at a Solution

Suppose that ##H \cup K \le G##. For contradiction, suppose that neither H nor K is a subset of the other. This implies that there exists an ##h \in H## that is not in ##K## and a ##k \in K## that is not in ##H##. Since ##H## and ##K## are individually subgroups, ##h^{-1} \in H## and ##k^{-1} \in K##. Now since ##h \in H \cup K## and ##k \in H \cup K##, we know that ##hk \in H \cup K##, since we supposed it's a subgroup.
I wouldn't say suppose here as it sounds as if it could be wrong. It is our given condition, i.e. we accept it as given truth. "... since ##H \cup K## is a subgroup." would be better.
We have three cases to check: If ##hk \in (H \cap K)##, then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - H## then ##(hk)k^{-1} = h \in K##, a contradiction. If ##hk \in (H \cup K) - K## then ##h^{-1}(hk) = k \in H##, a contradiction.
I think you don't need the cases. Just look at ##hk##. It's either in ##H## in which case ##k## is as well, or vice versa for the other possibility. You can say w.l.o.g. ##hk \in H## (the problem is symmetric in ##H## and ##K##) and have only one case to consider.

Mr Davis 97

## 1. What is a subgroup?

A subgroup is a subset of a larger group that itself forms a group. In other words, it contains all the necessary elements and operations to be considered a group on its own.

## 2. How is a subgroup related to the original group?

A subgroup is a smaller version of the original group, with the same operations and structure. It is a subset of the original group that maintains the same properties and follows the same rules.

## 3. What is the significance of the union of two subgroups?

The union of two subgroups, in this case H and K, forms a larger subgroup that contains all the elements and operations of both H and K. This helps to simplify and streamline the structure of the overall group.

## 4. What conditions must be met for H and K to form a subgroup?

In order for H and K to form a subgroup, they must both be subgroups of the same larger group and they must have at least one common element. In addition, the operations of H and K must be closed under the larger group's operations.

## 5. How can the conditions on H and K be determined?

The conditions on H and K can be determined by checking if they fulfill the requirements for being a subgroup, as mentioned in the previous question. This includes checking for closure, associativity, identity, and inverses for both H and K within the larger group.

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