Is this an equivalent definition of a normal subgroup?

Question

Let $G$ be a group and $N$ a subgroup. Consider the condition $$(\forall g\in G)(\exists x,y\in G)\ NgN=xNy.\tag1$$

If $N\lhd G$, then for each $g\in G$ we have $NgN=gNN=gN=gN\cdot1$, so the condition is satisfied.

Suppose $G$ is finite and condition $(1)$ is satisfied. Then by this, we have, for any $g\in G$, $$|N|=|xNy|=|NgN|=\frac{|N|^2}{|N\cap gNg^{-1}|},$$ and therefore $$|N\cap gNg^{-1}|=|N|.$$ Since $N\cap gNg^{-1}\subseteq N$ and $N$ is finite, we have that $N=gNg^{-1}$ for all $g\in G.$ This means that $N$ is normal in $G$.

Is this true in general that condition $(1)$ is equivalent to $N$ being normal in $G?$ There are three similar conditions: $$(\forall g\in G)(\exists x\in G)\ NgN=xN.\tag2$$ $$(\forall g\in G)(\exists y\in G)\ NgN=Ny.\tag3$$ $$(\forall g\in G)(((\exists x\in G)\ NgN=xN)\vee(\exists y\in G)\ NgN=Ny)).\tag4$$

Perhaps one of them works? (The first one is the one I first encountered, more or less naturally, which is why I used it. But it is clear that the other conditions work just as well for what I did. I don't know what happens for infinite groups.) $(1)$ is the weakest of the conditions, so if it implies that $N$ is normal, the others do too and, in consequence, all are equivalent. So to show that $(1)$ doesn't imply that $N$ is normal, it's enough to show that $(1)$ doesn't imply one of the other conditions. Please correct me if this is wrong.

As a general note, even $gNg^{-1}\subseteq N\implies gNg^{-1}=N$ fails for infinite groups (which I found surprising at first). — anon, Apr 13 '13 at 15:55
@Martin I don't know whether it does. I think I'm saying that (2) implies (1). Is there something wrong with what I said? — Bartek, Apr 13 '13 at 17:50
@Martin (2) implies (1), so if (1) implies normality, then (2) implies normality too, by the trasitivity of implication. — Bartek, Apr 13 '13 at 17:54
@Martin But normality implies all of them which gives what I said after "in consequence". — Bartek, Apr 13 '13 at 17:55
An implication employed in a deleted answer, $NgN=xN\implies Ng=xN$, is false in general. In particular, if $g^{-1}Ng\subsetneq N$, a possibility I mentioned in my original comment above (see this thread and links therein for explicit examples), and $x=g$, then $NgN=xN$ but $Ng\ne xN$. — anon, Apr 13 '13 at 18:29
Observe that condition (1) is equivalent to: For all $g\in G$ there exists $x\in G$ such that $NgNg^{-1}=xNx^{-1}$. — Diego, Apr 13 '13 at 18:43

Derek Holt · Accepted Answer · 2013-04-13T22:26:38.103

6

I believe there is an example of a non-normal subgroup satisfying (1) (and also (4)).

Let $G = \langle x,y \mid y^{-1}xy = x^2 \rangle$ and $N = \langle x \rangle$. Then $y^{-1}Ny < N$ but $yNy^{-1} \not< N$. (This group can also be defined as a group of $2 \times 2$ rational matrices.)

So, for all $k > 0$, $Ny^k < y^kN$ and $y^{-k}N < Ny^{-k}$ and hence $Ny^kN = y^kN$ and $Ny^{-k}N = Ny^{-k}$, which proves (1) and (4) when $g$ is a power of $y$.

The normal closure $K$ of $x$ in $G$ is an infinitely generated abelian group generated by elements $x_n = y^{-n}xy^n$ for $n \in \mathbb{Z}$, where $x_0=x$ and $x_n^2=x_{n+1}$. An arbitrary element $g \in G$ can be written as $y^kx_n^j$ for some $k,n,j$, or alternatively as $x_n^j y^k$ (with the same $k$ and $j$ but a different $n$). Since $K$ is abelian, each $x_n$ commutes with $N$, and so we get Condition (4) for all $g \in G$.

edited Apr 13 '13 at 22:26

answered Apr 13 '13 at 19:31

Derek Holt

90,008

1

As I understood, instead of "$\mid y^{-1}xy = y^2$" should be "$\mid y^{-1}xy = x^2$". – Alex Ravsky Apr 13 '13 at 19:59
I think this is related to the example you gave in the question about finding a group in which there was no simultaneous right and left transversal for a subgroup. – Geoff Robinson Apr 13 '13 at 20:08
@Derek Holt: One of us should be wrong. :-)
Why the implication after “hence” holds in here? : “So, for all $k > 0$, $Ny^k < y^kN$ and $y^{-k}N < Ny^{-k}$ and hence $Ny^kN = y^kN$”
– Alex Ravsky Apr 13 '13 at 20:22
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@Alex Ravsky: Yes it should have been $y^{-1}xy=x^2$ - I've corrected it. $Ny^k < y^kN$ implies $y^kN \le Ny^kN \le y^kNN=y^kN$. Why do you say that one of us should be wrong? I agree that (2) is equivalent to normality of $N$. – Derek Holt Apr 13 '13 at 22:30
@Derek Holt: "> Why do you say that one of us should be wrong?" Opps. I read "$\vee$" as "$\wedge$" in the formulation of condition (4). I’m sorry. – Alex Ravsky Apr 14 '13 at 05:21
@DerekHolt Are (1) and (4) are equivalent? (This is not terribly important, but I don't see it, and I think that if they are then it's non-obvious.) – Bartek Jan 01 '15 at 03:52
@GeoffRobinson I'm adding a link to that comment: http://math.stackexchange.com/questions/268219/mutual-set-of-representatives-for-left-and-right-cosets-what-about-infinite-gro#comment586218_268274 – Bartek Jan 01 '15 at 03:52

Alex Ravsky · Answer 2 · 2013-04-13T18:34:29.787

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@Bartek, Geoff Robinson: Thanks for you remarks. I agree with them. My idea failed. But I have another and I corrected my text accordingly to it.:

It seems that you are right, and the condition (2) is equivalent to the normality of the group $N$. The necessity is obvious. Now suppose that the group $N$ satisfies condition (2). The set $NgN$ contains a left coset $gN$ and is equal to the left coset $xN$. Thus $NgN=gN=xN$. Then $g^{-1}NgN=g^{-1}gN=N$. This implies that $g^{-1}Ng\subset N$ for each $g\in G$, and, therefore the group $N$ is normal.

edited Apr 13 '13 at 18:34

answered Apr 13 '13 at 16:59

Alex Ravsky

90,434

4

I see that $Ng=xN$ implies $NgN=xN$, but why the converse (which is the implication you're using)? – Bartek Apr 13 '13 at 17:11

Is this an equivalent definition of a normal subgroup?

2 Answers2