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Question:
Let $(G;.)$ a finite no commutative group. Find that an upper bound that to the probability that two element $X,Y$ of $G$ randomly chosen with an uniforme probability (that means the probability to pick one specific element of $G$ is $1/|G| $) commute is $\leq 5/8$

Answer:
I/Let's define the following sub set of $G$:
$S_1=\left \{ x \in G \; s.t. \; \forall y \in G \Rightarrow x.y=y.x\right \}$ = the set of all the elements of $G$ that commute with all the other elements of $G$
$S_{2 (x)} = \left \{ y \in G \; s.t. \; x.y=y.x\right \}$ = for a given $x$, $S_{2(x)}$ is the set of all the element $y\in G$ that commutes with $x$. Please remark that this set depends of $x$. This two set are sub group of $G$ (will not writte the proof here but it's by definition).

II/Moreover we should note that it exists at least one element $x' \in G$ that does not commute with all the element of $G$ and so $x' \notin S_1$ (if such element doesn't exist $G$ will be a comutative group). More than that all the element in $S_1$ commute with $x'$ by definition. So they are all in $S_{2(x')} \Rightarrow S_1 \subset S_{2(x')} \subset G$.
Now as $S_{2(x')} \subset G$ and $S_{2(x')}$ is too a group $\Rightarrow |S_{2(x')}|\leq |G|/2$.
Same argument for $S_1$ (which is a sub group too) $\Rightarrow |S_1|\leq|S_{2(x')}|/2\Rightarrow|S_1|\leq |G| /4$

III/ Now let majorate the probability:
$\mathbb{P}(X.Y=Y.X)=\mathbb{P}( \left \{X \in S_1 \cup Y \in S_1 \right \} \cup \left \{Y\in S_{2(x)} \cup X\in S_{2(y)} \right \} )\leq ????$
explanation of the event: X and Y commutes= if the X we picked commutes with all the other elements of G OR if the Y we picked commutes with all the other elements of G OR if the Y we picked belong to all the element in G that commutes with the X we picked OR if the X we picked belong to all the element in G that commutes with the Y we picked

But here i am stuck. I ve tried different ways (exclusion principle, re writte the probability via conditional probability... but maybe trying this i ve made mistakes) to majorate this probability (above) but i never succeed to get $5/8$. Can someone please help me to conclude?

Thank a lot.

PS: i ven't any special knowledge in algebra as i got this question in a probability book.

Arturo Magidin
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X0-user-0X
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1 Answers1

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You wrote everything you need from the algebraic part, except that you need to show that $|S_1| \leq |G|/4$ in more details - more specifically, that you can't have situation when $S_1 = S_{2(x')}$ for all $x'$ (or see this question to directly get $|S_1| \leq |G| / 4$).

The important part is that center is at most $1/4$ of the group, and elements not from center don't commutate with at least $1/2$ of the group.

We have $$P(x.y \neq y.x) \\= P(y \not \in S_2(x)) \geq P(y \not \in S_2(x) \wedge x \not \in S_1) \\ = P(x \not \in S_1) \cdot P(y \not \in S_{2(x)} | x \not \in S_1) \\ \geq \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$$

Thus $P(x.y = y.x) = 1 - P(x.y \neq y.x) \leq \frac{5}{8}$.

mihaild
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  • Thank for your answer (indeed i didn't try the complement event even if i tried conditional expectation). I don't know what is the center. And does my event (for the probability part) that i try to majorate is correct? – X0-user-0X Nov 29 '22 at 11:00
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    Center is set of elements that commutate with all others ($S_1$ in your notation). Yes, your event is correct, but I don't think such split (into intersecting events) will be helpful - the bound is tight, as in quaternion group we have exactly $40$ commutating element pairs out of $64$ total. – mihaild Nov 29 '22 at 11:52
  • @mihlaid How did you get the majoration if i am not wrong and by using de Morgan law + conditional probability formula, the best that i get is $3/4$ because: $\mathbb{P}(X.Y\neq Y.X)=\mathbb{P}((X.Y= Y.X)^{NOT})=\mathbb{P}( X \notin S_1 \cap Y \notin S_1 \cap Y \notin S_{2(x)} \cap X \notin S_{2(y)})=\mathbb{P}( X \notin S_1 \cap Y \notin S_1| Y \notin S_{2(x)} \cap X \notin S_{2(y)})\mathbb{P}(Y \notin S_{2(x)} \cap X \notin S_{2(y)})=\mathbb{P}(Y \notin S_{2(x)} \cap X \notin S_{2(y)}) \geq \frac{1}{4}$ (in an other way i get too 1/64 it's the most simple and worst bound to get) – X0-user-0X Nov 29 '22 at 12:11
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    How did you get your estimation of $P(Y \not \in S_{2(x)} \cap X \not \in S_{2(y)})$? And what is the first inequality in my answer you don't understand how to get? – mihaild Nov 29 '22 at 12:16
  • 1/How did you get: $P(x.y \neq y.x) \geq P(x \not \in S_1) \cdot P(y \not \in S_{2(x)} | x \not \in S_1)$? 2/ Becasue $\forall x \in G \Rightarrow S_1 \subseteq S_{2(x)}$ so if we know that Y not in $S_{2(x)}$ it is not in $S_1$ same for $X$ hence the probability of this conditional expectation is 1
    3/ I think i ve understood how you got it i will writte a more comment in 15 minuts (i hope it will be true) minutes now i ve to go     – X0-user-0X Nov 29 '22 at 12:22
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  • Added a bit more details, hopefully it's clearer now. 2. $x \not \in S_1$ doesn't imply that $y \not \in S_{2(x)}$ - fact that $x$ don't commute with everything doesn't mean it doesn't commute with $y$.
    • – mihaild Nov 29 '22 at 12:26
  • @mihlaid 1/thk i understand your answer now 2/ok but if we know that $Y \notin S_{2(x)}$ so $Y$ is not in $S_1$ for sure too right? because it means that Y does not commute with $X$ so $Y$ does not commute with all the element of $G$. So if i all ready know that X not in $S(y)_2$ and that Y not in $S(x) _2$ the probability that X and Y not in $S_1$ knowing those facts is equal to one right? – X0-user-0X Nov 29 '22 at 13:04
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    Yes, but in the conditional probability we are interested in other thing: we know that $x \not \in S_1$ and we want to know if $y \not \in S_{2(x)}$. – mihaild Nov 29 '22 at 13:08
  • 1/thk a lot foe your answer and time. 2/just to be sure that i ve understood so the way i get the lower bound 1/4 as explained in the previous comments is correct? (i mean without knowing the fact that there is an other bound wich is 3/8 because in this case it is obvious) – X0-user-0X Nov 29 '22 at 13:11
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    I don't understand, how you got this $1/4$. If it was as $P(y \not \in S_{2(x)} \cap x \not \in S_{2(y)} = P(y \not \in S_{2(x)}) \cdot P(x \not \in S_{2(y)}) = 1/2 \cdot 1/2$ then it's not correct: first, events are not independent, and second, you don't have $P(y \not \in S_{2(x)}) = 1/2$ (and it's in fact not true in general). – mihaild Nov 29 '22 at 13:17
  • not equal but as a a lower bound bjt it doesn't really matter as i ve understood your answer thk you – X0-user-0X Nov 29 '22 at 13:24
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    "commute", not "commutate" – Arturo Magidin Nov 29 '22 at 15:10