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So the problem is as follows:

Prove that if H is a subgroup of G, then H is a normal subgroup of G iff the following condition holds: $$\forall x,y \in G, xy \in H \iff yx \in H$$

It's all completed now:

We are given that H is a subgroup of G.

($\def\impl{\;\Rightarrow\;}\impl$) Assume H is a normal subgroup of G. So, $$\forall h \in H, \forall g \in G, ghg^-1 \in H.$$ Suppose $$\forall x,y \in G, xy \in H.$$ Since H is a normal subgroup of G and $$y \in G,$$ we know $$y(xy)y^-1\in H \impl yx(yy^-1)\in H \impl yxe \in H \impl yx \in H.$$ Similarly, suppose $$\forall x,y \in G, yx \in H.$$ Since H is a normal subgroup of G and $$x \in G,$$ we know $$x(yx)x^-1\in H\impl xy(xx^-1)\in H\impl xye \in H\impl xy \in H.$$

($\;\Leftarrow\;$) Assume $$\forall x,y\in G, xy\in H \iff yx \in H.$$ Now let $$a=yx \impl xa=x(yx) \impl (xa)x^-1=(xy)xx^-1 \impl xax^-1 = xy.$$ So $$ \forall x\in G, \forall a\in H, a \in H \impl xax^-1 \in H.$$ Therefore H is a normal subgroup of G.

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Katie
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1 Answers1

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Note first that the elements $xy$ and $yx$ are conjugate, as $$ x^{-1} (xy) x = y x. $$ Conversely, if $a$ and $b$ are conjugate, $b = z^{-1} a z$, say, then $$ b = (z^{-1} a) z, \quad a = z (z^{-1} a) $$ so $a = xy$ and $b = yx$, for $x = z$ and $y = z^{-1} a$.

Now use the definition of normal subgroup.

Andreas Caranti
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  • Which direction is this for? – Katie Mar 19 '13 at 09:48
  • @Katie, both directions. It shows that your statement is equivalent to the definition. – Andreas Caranti Mar 19 '13 at 09:54
  • Sorry, I'm having a hard time seeing how it implies anything. Using a similar idea, I was able to prove this in the forward direction, but I'm still stuck using this fact to prove it in reverse. Also, the definition of normal subgroup says that $$\forall h\in H, \forall g\in G, ghg^-1\in H.$$ This is distinct from if it were to say $$g^-1hg\in H$$ because we are not given that it is abelian. – Katie Mar 19 '13 at 10:01
  • @Katie, your assumption is $xy \in H$ implies $yx \in H$ for all $x, y \in G$. Write $a = xy$, and then $yx = x^{-1} a x$. So your assumption now reads $a \in H$ implies $x^{-1} a x \in H$ for all $a \in H$ and $x \in G$. This is exactly the definition of a normal subgroup $H$. – Andreas Caranti Mar 19 '13 at 10:24
  • Now I see what you're saying, thank you! – Katie Mar 19 '13 at 10:32