0

The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes.

Background information

Prove that if $\phi:G\rightarrow H$ is a group homomorphism with $\phi(G)=M\neq H,$ then $\phi$ is not an epimorphism in $\textbf{Grp}.$ [Hint: If $M$ has only 2 cosets in $H$, take $K=H/M,$ and define $\psi, \psi':H\rightarrow K$ by $\psi(h)=[h]$ and $\psi'(h)=e_K.$ If not, take $K$ to be the group of all permutations of $H.$ Choose $3$ different cosets $M, Mh', Mh''$ of $M$ in $H,$ and define $\sigma$ in $K$ by $\sigma(xh'')=xh', \sigma(xh')=xh''$ for $x\in M,$ while otherwise $\sigma(h)=h.$ Define $\psi,\psi':H\rightarrow K$ by $\psi(h)=$ left multiplication by $h$, and $\psi'(h)=\sigma^{-1}\psi(h)\sigma.$ Then in either case $\psi\phi=\psi'\phi$ but $\psi\neq \psi'.$]

Questions

I don't understand how $\psi \neq \psi'.$

According to the proof, suppose $h\in H\setminus M,$ then $\psi(xh'')=hxh'',$ and $\psi'(xh'')=(\sigma^{-1}\psi\sigma)(xh'')=\sigma^{-1}\psi\sigma(xh'')=\sigma^{-1}\psi(xh')=\sigma^{-1}(hxh'')$

But does $\sigma^{-1}(hxh'')=\sigma^{-1}(h)\sigma^{-1}(xh'')?$ (not sure if this is correct or how to justify it?) So that $\sigma^{-1}(h)=h$ and $\sigma^{-1}(xh'')=xh'.$ Hence $\sigma^{-1}(hxh'')=hxh'.$ Hence $\psi'(xh'')\neq \psi(xh'').$

Am I understanding the proof correctly? Thank you in advance

Shaun
  • 44,997
Seth
  • 3,325
  • 2
    I can't make any sense of your explanation. $\psi(xh'')$ is not equal to $xh''$, it is the permutation of $H$ defined by left multiplication by $xh''$. To show that $\psi \ne \psi''$, note that $\psi(h')(e_H) = h'$, but $\psi'(h')(e_H) = h''$. – Derek Holt Jun 16 '23 at 10:55
  • @DerekHolt I corrected the typo. It should be $\psi(xh'')=hxh''.$. Can I ask if $\sigma^{-1}(hxh'')=\sigma^{-1}(h)\sigma^{-1}(xh'')?$ If so, do I use the notion of group action to justify it. – Seth Jun 16 '23 at 11:29
  • 1
    $\psi(xh'') = hxh''$ still makes no sense. $\psi(xh'')$ is not an element of $H$, it is a permutation of the underlying set of $H$. Also $\sigma$ is not a group homomorphism,it too is a permutation of the underlying set of $H$. – Derek Holt Jun 16 '23 at 12:16
  • @DerekHolt it says $\psi(h)=$ left multiplication by $h$ means $\psi(x)=hx,$ so $\psi(xh'')=hxh''$ Also, why is it that $\psi$ is a permutation of the underlying set of $H$? I thought only $\sigma$ denotes permutation. I am trying to understand the hint and I asked couple of math grad students, they were not able to make sense of it. I also posted the contrapositive version of this statement from another post, apparently Arthuro from the comment in that post told me I won't have a chance in hell in understanding the proof until I understand group actions. – Seth Jun 16 '23 at 12:25
  • A permutation is a special kind of function, namely a bijection from a set to itself. The set can be any set. Also, the symbol you use to express a permutation can be any symbol; you are not restricted to the symbol $\sigma$. – Lee Mosher Jun 16 '23 at 12:29
  • @LeeMosher the contrapositive statement I posted here. For both proofs, I asked two math grad students about it, they both were equally confused. I know I am supposed to fill in the gaps, but I am not able to, and they were not able to either, that is why I am asking what is confusing me for both on here. – Seth Jun 16 '23 at 12:35
  • $\psi(h) = $ left multiplication by $h$ does not mean $\psi(x)=hx$. It means that $\psi(h)$ is the permutation of $H$ defined by left multiplication by $h$. In other words $\psi(h)(x) = hx$ for all $x \in H$. – Derek Holt Jun 16 '23 at 12:54
  • @DerekHolt how does $\psi'(h)=\sigma^{-1}\psi(h)\sigma?$ and from there $\psi\neq \psi'?$ – Seth Jun 16 '23 at 12:59
  • $\psi'(h) = \sigma^{-1} \psi(h) \sigma$ is the definition of $\psi'$. I will answer the question later. – Derek Holt Jun 16 '23 at 13:23

1 Answers1

2

By definition $\psi(h)$ is a permutation of $H$ defined by $\psi(h)(x) = hx$ for $x \in H$.

$\sigma$ is also defined as a permutation of $H$, and $\psi'(h) = \sigma^{-1}\psi(h)\sigma$ is defined to be the composite of the three permutations $\sigma^{-1},\,\psi(h),\,\sigma$.

I claim that $\psi'(h') \ne \psi(h')$, which proves that $\psi \ne \psi'$. To prove this, we have to show that $\psi(h')$ and $\psi'(h')$ are different permutations of $H$, and to do that it is enough to show that they have different actions on some point of $H$.

Let's consider their actions on the identity element $e_H$ of $H$.

We have $\psi(h')(e_H) = h'$, and $$\psi'(h')(e_H) = \sigma^{-1}(\psi(h')(\sigma(e_H))) = \sigma^{-1}(\psi(h')(e_H)) = \sigma^{-1}(h') = h'' \ne \psi(h')(e_H).$$ So $\psi(h') \ne \psi'(h')$ as claimed.

Derek Holt
  • 90,008
  • is it because $\psi'(h)$ being equal to left multiplication by $h$, that means it is consider to be a left regular representation, another words a group action? Then $h\cdot x=hx=\psi(h)(x),$ where $h\in H, x\in M.$ For the cosets, we also have $h\cdot M=hM,$ and $h\cdot x=hxh^{-1}$ But how do I relate to $\psi'(h)=\sigma^{-1}\psi(h)\sigma,$ I assume that $\sigma$ is also a left regular representation. I learned about left regular representation from Hungerford's undergraduate Abstract Algebra text. In there, he uses the notation: ... . – Seth Jun 17 '23 at 04:55
  • ...For all $a\in G$, $\phi_a(x)=ax.$ He also uses the notation $\phi(x)(a)$ to mean $\phi_a(x).$ I was reading up on group actions and trying to figure out which notation is what the whole day. My apologies for my late reply. – Seth Jun 17 '23 at 04:57
  • $\psi$ is the left regular representation of the group $H$. But $\sigma$ is not a representation at all, it is a single permutation. – Derek Holt Jun 17 '23 at 08:28
  • Sorry but I cannot make any sense of what you write, or what it has to do with my answer. I have answered the question. If there is some step that you do not understand then let me know. – Derek Holt Jun 17 '23 at 18:51
  • I want to know if $\psi'(h)=\sigma^{-1}\psi(h)\sigma$ is consider a left regular representation of the group $H$ from the way it is defined and also because after rearranging by conjugation, then $\psi(h)=\sigma\psi'(h)\sigma^{-1}.$ Sorry, I delete the comment since I made some LaTex error and it was not displaying my comment properly. – Seth Jun 17 '23 at 18:55
  • It's equivalent to the left regular representation. That is a technical term meaning exactly that it is defined from the regular representation by conjugating by a fixed permutation - which is $\sigma$ in this case. – Derek Holt Jun 18 '23 at 08:32
  • how did you figure out from the proof that $\sigma$ is not a regular representation. I thought every permutation group satisfies the two properties for regular representation. – Seth Jun 18 '23 at 09:26
  • As I said earlier, $\sigma$ is not a representation, and it is not a group, it is a single permutation.You seem to be very confused about what all these words mean! Your sentence "I thought every permutation group satisfies the two properties for regular representation" is just nonsense. I am sorry but I cannot answer any more questions unless you think there is a mistake in the proof. – Derek Holt Jun 18 '23 at 11:08
  • I think when I saw "take K to be the group of all permutations of H", I thought $\sigma$ is a permutation group. I just got confused. – Seth Jun 18 '23 at 15:42