Let H≤G Prove x−1y−1xy∈H∀x y∈G⇔H⊴G and GH is abelian.

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Melinda McCombs · Accepted Answer

Step 1  For the forward direction, you have that  y−1xy∈xH  for all x and y in G, as you say. Therefore, if y is an arbitrary element of G, and x is an arbitrary element of H, then xH=H, and therefore y−1xy∈H. This is exactly what it means for H to be a normal subgroup of G.  To see that GH is abelian, you need only show that xyH=yxH for any x,y∈G. This is equivalent to saying that x−1y−1xyH=H for all x and y, which is of course true by your hypothesis.  It is quite easy to get the reverse direction from the previous paragraph.

ramirezhereva · Accepted Answer

Step 1 For backward direction, you already started really well. Since GH is abelian, we have H=(xH)(x−1H)(yH)(y−1H)=(x−1H)(y−1H)(xH)(yH)=(x−1y−1xy)H since we can swap places of elements in GH. For forward direction, in order to show H⊴G, pick an element of the form g−1hg∈G. Then, h−1g−1hg∈H⇒g−1hg∈hH=H⇒g−1Hg⊂qH⇒⊴ I leave as an exercise to you to show that GH is abelian.

Let H\leq GProvex^{-1}y^{-1}xy\in H\forall xy\in G\Leftrightarrow H\unlhd Gand\frac{G}{H} is abelian.

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