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

agreseza

agreseza

Answered question

2022-01-12

Let HG
Prove
x1y1xyHx
yGHG
and
GH is abelian.

Answer & Explanation

Melinda McCombs

Melinda McCombs

Beginner2022-01-13Added 38 answers

Step 1
For the forward direction, you have that
y1xyxH
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 y1xyH. 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,yG. This is equivalent to saying that x1y1xyH=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

ramirezhereva

Beginner2022-01-14Added 28 answers

Step 1
For backward direction, you already started really well. Since GH is abelian, we have
H=(xH)(x1H)(yH)(y1H)=(x1H)(y1H)(xH)(yH)=(x1y1xy)H
since we can swap places of elements in GH.
For forward direction, in order to show HG, pick an element of the form g1hgG. Then,
h1g1hgHg1hghH=Hg1HgqH⇒⊴
I leave as an exercise to you to show that GH is abelian.

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