let $G$ be a non abélien finite group then the probability that two elements commute is less or equal to $\frac{5}{8}$ \
Proof:
Let $(\Omega, T, \mathbb{P})$ be a probability space $X$ and $Y$ be two random variables such that $X \sim U(G)$ and $Y \sim U(G)$ now consider the event $(XY=YX)={\omega \in \Omega X(\omega)Y(\omega)=Y(\omega)X(\omega)}$ then in that case $\mathbb{P}((XY=YX))=\frac{1}{|G|^{2}} |((XY=YX))|=\frac{1}{|G|^{2}} \sum_{(x,y) \in G^{2}, xy=yx} 1$ now for $x$ in $G$ let’s set $C_x={y \in G, xy=yx}$ and $Z(G)={x \in G, \forall y \in G, xy=yx }$ then $\mathbb{P}((XY=YX))= \frac{1}{|G|^{2}} \sum_{x \in G} \sum_{y \in \G} H(x,y)$ with $H(x,y)=1$ if $xy=yx$ and 0 else which equals to $\frac{1}{|G|^{2}} \sum_{x \in G} |C_x|$ Now for all x in $G$ $C_x$ is a subgroup of $G$ so using Lagrange’s theorem, $|C_x|$ divides $|G|$, now using this there are two possibilities, either, $x \in Z(G)$ so is in the centre of the group, in that case $|C_x|=|G|$, or $C_x \subset_{\neq} G$ in that case we can deduce that $|C_x| \le \frac{1}{2} |G|$ so $(Z(G), G/Z(G))$ forming a partition of G and the family $(|C_x|){x \in G}$ being summable we have $\sum{x \in G} |C_x|=\sum_{x \in Z(G)} |C_x| + \sum_{x \in G/Z(G)} |C_x| \le \sum_{x \in Z(G)} |G| + \frac{1}{2} \sum_{x \in G/Z(G)} |G| \le |Z(G)| |G| + \frac{1}{2}(|G|-|Z(G)|)|G| \le \frac{1}{2} (|G|^{2}+|Z(G)| |G|)$, finally we can prove that for $Z(G)| \le \frac{1}{4} |G|$ indeed let $x \in G/Z(G)$ then we have $C_x \subset_{\neq} G$ so $|C_x| \le \frac{1}{2} |G|$ but we also have $Z(G) \subset_{\neq} C_x$ so $|Z(G)| \le \frac{1}{2} |C_x| \le \frac{1}{4} |G|$ so finally $\frac{1}{2} (|G|^{2} + |Z(G)| |G|) \le \frac{1}{2}( (1+ \frac{1}{4}) |G|^{2} \le \frac{5}{8} |G|^{2}$ thus $\mathbb{P}(XY=YX) \le \frac{5}{8}$
#cool result mixing abstract algebra and probability
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I just wanted to share a cool result I found
On The probability that two elements of a non commutative finite group commute
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On The probability that two elements of a non commutative finite group commute
Really cool!
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Really cool!
It is! It shows that there is quite a difference between an abelian finite group an a non abelian one in terms of probability
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what is the source?
Suppose that there exists a prime number $p$ and an element $x ∈ G$ such that the cardinality of the conjugacy class of x is divisible by $p$. Find a good/sharp upper bound for $\bar{c}$(G).
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For $g\in G$, denote by $Z(g)$ the centralizer of $g$. That is, $Z(g)={x\in G\mid xg=gx}$. It is straightforward to see $Z(g)$ is a subgroup of $G$ for all $g$. Furthermore, denote by $Z(G)$ the center of $G$. That is, $Z(G)={x\in G\mid\forall g\in G,xg=gx}$. Again, we can check $Z(G)$ is a subgroup of $G$. Observe that $G$ is abelian if and only if $G=Z(G)$.
Now, some important facts:
- For all $g\in G$, $Z(G)\le Z(g)\le G$.
- The integer $|G|/|Z(G)|$ cannot be prime.
Proof. Suppose $x\in Z(G)$. Then, $xg=gx$, so $x\in Z(g)$. This shows the first subgroup inclusion. The second was stated above. Recall that the order of a subgroup divides the order of the group. Hence, the numbers $k=|G|/|Z(G)|,m=|G|/|Z(g)|,n=|Z(g)|/|Z(G)|$ are all integers. Furthermore, we clearly have $k=mn$. This holds for all choices of $g$. Suppose $k$ is prime. Then, $|Z(G)|<|G|$, so $G\ne Z(G)$ and $G$ is not abelian. Then, take $g\in G\setminus Z(G)$. Since $k$ is prime and $k=mn$, we must have either $m=k,n=1$ or $m=1,n=k$. In the first case, $n=1$ implies $|Z(g)|=|Z(G)|$, so $Z(g)=Z(G)$. But observe $g\in Z(g)$ and $g\notin Z(G)$. Hence, this is not possible. In the second case, $m=1$ implies $|G|=|Z(g)|$, so $G=Z(g)$ and everything commutes with $g$. But then, $g\in Z(G)$, which is not the case. Thus, $k$ cannot be prime as claimed.
We can finally proceed with the problem. Let $p(G)$ denote the probability that two (uniformly) randomly picked elements of the group commute. Since $g$ commutes with precisely those elements in $Z(g)$, it is straightforward to check that
[ p(G) = \sum_{g\in G}Z(g)/|G|^2. ]Suppose $G$ is not abelian. Then, $G\ne Z(G)$. Thus, $|G|/|Z(G)|>1$. Since it cannot be prime, $|G|/|Z(G)|\ge 4$. This means that $|Z(G)|\le |G|/4$. Then, if $g\in Z(G)$, we have $|Z(G)|=|G|$. However, if $g\notin Z(G)$, then $|Z(g)|<|G|$. Since $|Z(g)|$ divides $|G|$, then in fact $|Z(g)|\le|G|/2$. Thus, at least $3|G|/4$ elements have centralizer of size at most $|G|/2$ and all other elements have centralizer of size at most $|G|$. Hence, if $G$ is not abelian we have
\begin{align*}
p(G)&=\sum_{g\in G}Z(g)/|G|^2\
p(G)&\le\frac{(3|G|/4)\cdot(|G|/2)+(|G|/4)\cdot(|G|)}{|G|^2} \
p(G)&\le\frac34\cdot\frac12+\frac14\cdot1\
p(G)&\le\frac58,
\end{align*}as desired. $\blacksquare$
Remark. This bound is sharp, as can be seen from the dihedral group with $8$ elements. Indeed, if we present $D_8=\langle R,T\mid R^4,TR=R^3T\rangle$, then we see $|Z(1)|=|Z(R^2)|=8$ and $|Z(g)|=4$ for all other $g\in D_8$. Then, $p(D_8)=(8\cdot2+4\cdot6)/64=40/64=5/8$. Hence, we cannot do better than $5/8$ without additional assumptions.
