Let G be a group. Show that G≠qH∪KG=H∪K for any two subgroups H≤GH≤GandK≤GK≤G.

Let G is a group and H,K are two non trivial subgroup of G. If possible, let G=H∪K. Now Since, H not equal to G and K not equal to G, so we must have HH not equal to K and H and K can not be subset of each other. Therefore there exists h∈H∖Kandk∈K∖H. Now since G is a group and h,k∈H∪K=G, then hk∈G=H∪K. So, hk is either in H or in K. Suppose hk∈H, since H is a subgroup and h∈H, so h−1∈H.h−1∈H. Therefore, h−1(hk)=(h−1×h)k=k∈H, which is against our choice, contradiction. Similarly, we can show that hk can not be in K. Henc, $G $ not equal to H∪K, for non trivial subgroups H and K.

Find out the answer to "Let G be a group. Show that G≠qH∪KG=H∪K..."

High School
Algebra
Algebra I
Exponents and radicals

necessaryh

Answered question

2020-11-06

Let G be a group. Show that

G \neq q H \cup K G = H \cup K

for any two subgroups

H \leq G H \leq G and K \leq G K \leq G

Answer & Explanation

delilnaT

Skilled2020-11-07Added 94 answers

Let G is a group and H,K are two non trivial subgroup of G. If possible, let

G = H \cup K

. Now Since, H not equal to G and K not equal to G, so we must have HH not equal to K and H and K can not be subset of each other. Therefore there exists

h \in H ∖ K and k \in K ∖ H

. Now since G is a group and

h, k \in H \cup K = G

, then

h k \in G = H \cup K

. So, hk is either in H or in K. Suppose

h k \in H

, since H is a subgroup and h∈H, so

h - 1 \in H . h^{- 1} \in H

. Therefore,

h^{- 1} (h k) = (h^{- 1} \times h) k = k \in H

, which is against our choice, contradiction. Similarly, we can show that hk can not be in K. Henc, $G $ not equal to

H \cup K

, for non trivial subgroups H and K.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra I

Didn't find what you were looking for?