Showing a subgroup is a normal iff its left and right cosets (with distinct elements) are not disjoint and are equal

Question

I am learning group theory and have come across the following problem statement

Show that $K$ is a normal subgroup of a group $G$ if and only if $$\forall_{x,y\in G} xK\cap Ky \neq \emptyset \implies xK=Ky $$ I am having a lot of trouble showing this. I give approach below:

Approach

$\Rightarrow$: $K$ a normal subgroup means that $\forall_{g\in G}gKg^{-1}=K$. Define by ${}_K|\in\text{Rel}_2(G)$ the right $K$ divisor relation, $_{K}|(x,y):= Kx\ni y$ and similarly the left $K$ divisor relation, $|_K(x,y)=xK\ni y$. Then $K$ a normal subgroup implies $g\hspace{0.1cm}{}_K|g^{-1}\land g|_Kg^{-1}$, or, $Kg\ni g^{-1}$ and $gK\ni g^{-1}$. Similarly, $g^{-1}{}_K|g\land g^{-1}|_K g$, or $Kg^{-1}\ni g$ and $g^{-1}K\ni g$. This implies that $g^{-1}\in gK\cap Kg\neq \emptyset$ as well as $g\in g^{-1}K\cap Kg^{-1}$. Since this holds for all $g\in G$, it follows that $gK=Kg$. If we then just replace by $g$ and $g^{-1}$, $x$ and $y$, then the forward implication is proven.

$\Leftarrow$: Suppose $gK\cap Kg\neq \emptyset\implies gK=Kg$. Then there exists an element $c\in gK\cap Kg$ such that for all elements $k\in K$, $gk=c=kg$, which implies that $gkg^{-1}$ and $g^{-1}kg$, which implies that $gKg^{-1}$ which defines a normal subgroup.

Concerns, Confusion, and Comments

My first concern arises by saying that $K$ a normal subgroup implies what I wrote out with the left and right $K$ divisor relations. It is intuitive that this would follow, but without stating the point of the problem. I imagine I need to state why such a thing is implied. Secondly, I have chosen to do this for an arbitrary element $g\in G$ rather than with the two distinct values $x$ and $y$ as implied by the problem statement. I imagine it cannot simply be assumed that $g=x$ and $g^{-1}=y$. I would appreciate any tips on how to solve this problem. Thanks!

Note: I did not come up with the left and right divisor relations. These are coset relations my professor introduced these concepts to us with.

Answer 1

The proof of a direct statement is unnecessarily verbose. The proof can be made quite short:

If $K$ is a normal subgroup, then $Kx=xK$ for any $x\in G$. Therefore $$ xK\cap Ky\neq\varnothing\Rightarrow Kx\cap Ky\neq\varnothing\Rightarrow Kx=Ky\Rightarrow xK=Ky. $$

The proof of the converse statement is incorrect. This proof is even shorter:

Suppose the condition you mentioned is satisfied. You need to prove that $xK=Kx$ for any $x\in G$. Since $xK\cap Kx\neq\varnothing$ (why?) it follows from the condition that $xK=Kx$.

Answer 2

I will go line-by-line of your answer, because there is a lot of confusion and incorrect statements.

The way you introduce left and right division is a problem. You write "$|_K(x,y)$", but then use the notation as infix, $x|_Ky$; that is confusing. You have for a subgroup $K$ of a group $G$, and $x,y\in G$, $$\begin{align*} x\;|_K\; y &\iff y\in xK\\ x\;{}_K|\;y&\iff y\in Kx \end{align*}$$ Fair enough as far as that goes.

Then $K$ a normal subgroup implies $g\;{}_K|\; g^{-1}\wedge g\;|_K g^{-1}$

This is false. The fact that $K$ is normal implies that $K=gKg^{-1}$, not that $g^{-1}\in Kg$ or that $g^{-1}\in gK$. In fact:

Prop. Let $G$ be a group, $K$ a subgroup. Then $g^{-1}\in Kg$ if and only if $g^2\in K$; likewise, $g^{-1}\in gK$ if and only if $g^2\in K$.

Proof. If $g^{-1}\in Kg$ then there exists $x\in K$ such that $g^{-1}=xg$. Multiplying on the right by $g^{-1}$ we get $g^{-2}=x$ and taking inverses we get $g^2 = x^{-1}\in K$. Conversely, if $g^2\in K$, then let $k\in K$ be such that $g^2=k$. Then $g^{-2}=k^{-1}$, and multiplying by $g$ on the right we get $g^{-1}=k^{-1}g\in Kg$. A symmetric argument holds for $gK$. $\Box$

Similarly, $g^{-1}\;{}_K|\;g \wedge g^{-1}\;|_K\;g$.

For similar reasons, this is false in general.

You seem to have misunderstood what $gKg^{-1}=K$ means, or else you have somehow incorrectly derived that $gKg^{-1}=K$ implies $g^{-1}\in gK$ (or $Kg$). That is just not true.

This implies that $g^{-1}\in Kg\cap gK\neq\varnothing$ as well as $g\in Kg^{-1}\cap g^{-1}K$.

Well, this would follow from what you had, but what you had is false. Inn general, again, you do not have $g^{-1}\in Kg\cap gK$. What you do have, for sure, though, is $g\in Kg\cap gK$. Why? Because $Kg=\{xg\mid x\in K\}$, and $gK=\{gy\mid y\in K\}$, and taking $x=y=e$ you get $g\in Kg\cap gK$.

Since this holds for all $g\in G$, it follows that $gK=Kg$.

This does not follow from what you had. Remember what you are trying to prove: you are trying to prove that for any $x,y\in G$, if $xK\cap Ky\neq\varnothing$, then $xK=Ky$. Not only have you not shown this for arbtirary $x$ and $y$, you didn't really derive it for $g$, either. All you did was argue that $gK$ and $Kg$ have nontrivial intersection. But that does not mean that they are equal... at least, you have not shown why that would follow. The fact that this holds for all $g$ doesn't give the equality of $gK$ and $Kg$.

If we then just replace $g$ and $g^{-1}$, [with] $x$ and $y$, then the forward implication is proven.

That does not work. Remember that you started with $gKg^{-1}=K$. That property follows from $K$ being normal. If you replace $g$ with $x$ and $g^{-1}$ with $y$, you would need $xKy=K$ to hold. Does it? Not necessarily. You can freely replace $g$ with $x$... but then $g^{-1}$ needs to be replaced with $x^{-1}$, not with another arbitrary element. Because $g^{-1}$ depends on $g$, so its replacement needs to depend on the replacement of $y$.

On to the converse.

Suppose that $gK\cap Kg\neq\varnothing\implies gK=Kg$.

That's fine. It's a consequence of the condition you are assuming, substituting $g$ for $x$ and $g$ for $y$ (you can replace $x$ and $y$ for the same thing or related things because $x$ and $y$ are not assumed to have any relation; what you cannot do is replace things that are related in some specific way by arbitrary $x$ and $y$).

Then there exists an element $c\in gK\cap Kg$...

Ehr... no. The thing you are assuming is that if there is a $c\in gK\cap Kg$, then $gK=Kg$. But you are not assuming that there is a $c\in gK\cap Kg$. You would have to show there is one. (In fact, there is: $g = ge\in gK$, and $g=eg\in Kg$, since $e\in K$).

Then there exists an element $c\in gK\cap Kg$ such that for all elements $k\in K$, $gk=c=kg$,

This is just plain wrong. If $c\in gK\cap Kg$, then there exist elements $k_1,k_2\in K$ such that $c=gk_1$ and $c=k_2g$. Not necessarily the same element, and not "for all" elements. If it were for all elements, then you are saying that $K$ has just one element! Because if $k,k'\in K$, then $gk=c=gk'$, so $k=k'$. So this assertion cannot possibly be correct (unless $K$ is trivial...)

which implies that $gkg^{-1}$ and $g^{-1}kg$,

Which implies what about them? The subject of this sentence is "$gkg^{-1}$ and $g^{-1}kg$". But you are not saying what it implies about them.

which implies that $gKg^{-1}$

Again, what does it imply about $gKg^{-1}$?

which defines a normal subgroup.

No. Your definition of normal subgroup is that $gKg^{-1}$ is equal to $K$ for all $g$. Here you just have $gKg^{-1}$, naked and with nothing to do and nothing said about it. The whole of the argument in this part is the following run-on sentence:

Then there exists an element $c\in gK\cap Kg$ such that for all elements $k\in K$, $gk=c=kg$, which implies that $gkg^{-1}$ and $g^{-1}kg$, which implies that $gKg^{-1}$ which defines a normal subgroup.

That's a word salad, but is unintelligible. And certainly incorrect.

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Now the proof:

$\implies)$ You assume $K$ is normal, so $gKg^{-1}=K$ for all $g\in G$. You want to prove that for all $x$ and $y$ in $G$, if $xK\cap Ky\neq\varnothing$, then $xK=Ky$. To that end, let $x$ and $y$ be elements of $G$ such that $xK\cap Ky\neq\varnothing$.

We want to show that $xK=Ky$.

If $gKg^{-1}=K$ for all $g$, then multiplying on the right by $g$ you get that $gK=Kg$ for all $g$. In particular, $Ky=yK$.

So we have that $xK\cap Ky = xK\cap yK\neq\varnothing$. But now $xK$ and $yK$ are both left cosets, and we know that two left cosets are either disjoint or identical. Since $xK$ and $yK$ are not disjoint, they are identical. And since $yK=Ky$, we have $xK=yK=Ky$, proving that $xK=Ky$, which is what we wanted to prove.

$\Leftarrow)$ Suppose that for any $x,y\in G$, if $xK\cap Ky\neq\varnothing$, then $xK=Ky$. We want to prove that for all $g\in G$, $gKg^{-1}=K$. To that end, let $g\in G$ be arbitrary.

Note that $g\in gK = \{gk\mid k\in K\}$ by taking $k=e$; and that $g\in Kg = \{kg\mid k\in K\}$. Therefore, $gK\cap Kg\neq\varnothing$. By our assumption, this implies that $gK=Kg$.

Now multiplying on the right by $g^{-1}$ we get $gKg^{-1}=K$, which is what we wanted to prove. $\Box$