Answer to Question #314565 in Abstract Algebra for Azooz

Question

Question

For sets H and K, we define the intersection H n K by H n K = {xI

X E H and X € K%. Show that if H ≤ G and K < G, then H n K < G.

(Remember: ≤ denotes "is a subgroup of," not "is a subset of.")

Answer 1

Let GG be a group with identity ee and let H\leq GH≤G and K\leq GK≤G.

By definition we have that H\cap K\subseteq GH∩K⊆G.

Since H\leq GH≤G and K\leq GK≤G we have that e\in He∈H and e\in Ke∈K which gives us that e\in H\cap Ke∈H∩K.

Given x,y\in H\cap Kx,y∈H∩K we have that x,y\in Hx,y∈H and x,y\in Kx,y∈K and then since H\leq GH≤G and K\leq GK≤G we have that xy\in Hxy∈H and xy\in Kxy∈K which gives us that xy\in H\cap Kxy∈H∩K. Thus H\cap KH∩K is closed under the operation of GG.

Given x\in H\cap Kx∈H∩K we have that x\in Hx∈H and x\in Kx∈K and then since H\leq GH≤G and K\leq GK≤G we have that x^{-1}\in Hx

−1

∈H and x^{-1}\in Kx

−1

∈K which gives us that x^{-1}\in H\cap Kx

−1

∈H∩K.

Hence H\cap K\leq GH∩K≤G.