Mysterious Characterization of $A_5$ inside $S_5$

Solution 1:

To sketch a partial solution to this we puzzle we can use some group character theory. Below I have included the character tables of $Sym(5)$ and $Alt(5)$. We first observe that the function

$$\chi(g)=|fix(g)|$$

is a group character, and by a well known theorem (Theorem 2.6.1) we see that $\chi(g)=\rho_1(g)+\rho_4(g)$, restricting to $Alt(5)$ we see that $\chi(g)=\phi_1(g)+\phi_4(g)$.

Now fix $\sigma=(12345)$ and pick and conjugacy class representatives in $Alt(5)$:

$\pi_1=id$,

$\pi_2=(12)(34)$,

$\pi_3=\sigma$

and $\pi_4=\sigma^{-1}$.

We have:

$[\sigma,\pi_1]=id$,

$[\sigma,\pi_2]=(12453)$,

$[\sigma,\pi_3]=(134)$,

$[\sigma,\pi_4]=id$

and $[\sigma,\pi_5]=(15432)=\sigma^{-1}$.

Moreover,

$\sigma^2[\sigma,\pi_1]=\sigma^2$,

$\sigma^2[\sigma,\pi_2]=(142)$,

$\sigma^2[\sigma,\pi_3]=(15243)$,

$\sigma^2[\sigma,\pi_4]=\sigma^2$

and $\sigma^2[\sigma,\pi_5]=(12345)=\sigma$.

Now, evaluating the character

$\phi_1([\sigma,\pi_i])+\phi_4([\sigma,\pi_i])+\phi_1(\sigma^2[\sigma,\pi_i])+\phi_4(\sigma^2[\sigma,\pi_i])=\chi([\sigma,\pi_i])+\chi(\sigma^2[\sigma,\pi_i])$

on those conjugacy classes gives either $2$ or $5$, and we're done.

For example, for $\pi_2$ we have

$\chi(\sigma^2[\sigma,\pi_i])=\phi_1([\sigma,\pi_2])+\phi_4([\sigma,\pi_2])+\phi_1(\sigma^2[\sigma,\pi_2])+\phi_4(\sigma^2[\sigma,\pi_2])=\phi_1((12453))+\phi_4((12453))+\phi_1((142))+\phi_4((142))=1-1+1+1=2$.

-edit-

I'll include some remarks about why this not a complete solution and what remains to be proven. The key mystery seems to be that to obtain $2$ or $5$ we require the elements $[\sigma,\pi]$ and $\sigma^2[\sigma,\pi]$ to have orders $(1,5)$, $(1,3)$, $(5,1)$, $(3,1)$ or $(2,2)$. Why this happens, I do not know. That being said, the fact that the commutator of two elements of $Alt(5)$ is clearly going to be contained in $Alt(5)$, also multiplying a commutator of $Alt(5)$ elements by an element of $Alt(5)$ will also be in $Alt(5)$.

The character table of $Sym(5)$:

$$ \begin{array}{c|rrrrrrr} \rm class&\rm1&\rm2&\rm2^2&\rm3^1&\rm4^1&\rm5^1&\rm6^1\cr \rm size&1&10&15&20&30&24&20\cr \hline \rho_{1}&1&1&1&1&1&1&1\cr \rho_{2}&1&-1&1&1&-1&1&-1\cr \rho_{3}&4&-2&0&1&0&-1&1\cr \rho_{4}&4&2&0&1&0&-1&-1\cr \rho_{5}&5&1&1&-1&-1&0&1\cr \rho_{6}&5&-1&1&-1&1&0&-1\cr \rho_{7}&6&0&-2&0&0&1&0\cr \end{array} $$

The character table of $Alt(5)$:

$$ \begin{array}{c|rrrrr} \rm class&\rm1&\rm2^2&\rm3^1&\rm5_A&\rm5_B\cr \rm size&1&15&20&12&12\cr \hline \phi_{1}&1&1&1&1&1\cr \phi_{2}&3&-1&0&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}\cr \phi_{3}&3&-1&0&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}\cr \phi_{4}&4&0&1&-1&-1\cr \phi_{5}&5&1&-1&0&0\cr \end{array} $$