Timeline for Are there always more conjugacy classes in the kernel of a morphism to $Z_2$ than not?
Current License: CC BY-SA 4.0
12 events
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| May 28, 2021 at 21:12 | comment | added | LSpice | I agree that describing them as "$G$-conjugacy classes which stay conjugacy classes for $H$" or "$H$-conjugacy classes which stay conjugacy classes for $G$" are both valid, and equivalent; I was just confused by "$G$-conjugacy classes in $H$ which stay conjugacy classes in $G$". | |
| May 28, 2021 at 21:09 | comment | added | David E Speyer | We want subsets of $H$ which are conjugacy classes for both $G$ and for $H$. We can either think of them as $G$-conjugacy classes which stay conjugacy classes for $H$, or as $H$-conjugacy classes which stay conjugacy classes for $G$. | |
| May 28, 2021 at 18:57 | comment | added | LSpice | Should be $H$-conjugacy classes that stay conjugacy classes in $G$, right? \\ A silly version of your desired action (suitable modification works for $\mathbb Z/p\mathbb Z$): multiply by a fixed element of $Z(G) \setminus H$, if one exists. | |
| May 28, 2021 at 18:54 | comment | added | David E Speyer | @LSpice Yes, I agree with your generalization to other primes. More specifically, I think it shows that the number of $G$-conjugacy classes in $H$ which stay conjugacy classes in $G$ is equal to the number of $G$-conjugacy classes lying over any fixed nonzero $a \in \mathbb{Z}/p \mathbb{Z}$. | |
| May 28, 2021 at 18:10 | comment | added | LSpice | By the way, the reason I asked for clarification about $H$ is that the question also uses $H$, with a different meaning. Speaking of that, wouldn't the same argument, with any target $\mathbb Z/p\mathbb Z$ with $p$ prime, show that $(p - 1)\cdot\#\text{conj. classes in $H$} \ge \#\text{conj. classes not in $H$}$, thus answering one possible version of the generalised question? | |
| May 28, 2021 at 12:35 | comment | added | David E Speyer | I suppose that the natural question this argument raises is to find an action of $\mathbb{Z}/2 \mathbb{Z}$ on $\mathrm{Conj}(G)$ with $a$ orbits of size $2$ and $b$ orbits of size $1$. (Since we have actions of $\mathbb{Z}/2 \mathbb{Z}$ on $\mathrm{Conj}(H)$, $\mathrm{Irrep}(H)$ and $\mathrm{Irrep}(G)$.) | |
| May 28, 2021 at 7:21 | vote | accept | Clark Lyons | ||
| May 29, 2021 at 4:06 | |||||
| May 28, 2021 at 6:56 | vote | accept | Clark Lyons | ||
| May 28, 2021 at 6:56 | |||||
| May 28, 2021 at 3:34 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| May 28, 2021 at 3:33 | comment | added | LSpice | $\DeclareMathOperator\Hom{Hom}$Proof of the well known fact: $\Hom_H(V_H, V_H) = \Hom_G(V_H^G, V) = \Hom_G(V \oplus (V \otimes \epsilon), V) = 1 + [V \otimes \epsilon \cong V]$. | |
| May 28, 2021 at 3:29 | comment | added | LSpice | $H = \ker \phi$? (I guess this comes later, when you define $\epsilon$; so I guess you just want to start with $H$ any index-2 subgroup?) | |
| May 28, 2021 at 3:27 | history | answered | David E Speyer | CC BY-SA 4.0 |