1

I am stuck on the following problem from these notes (problem 2.23(b)).

Let $G$ be a group, and let $H ⊂ G$ be a subgroup with $H \not = G$. Construct a group $K$ and two group homomorphisms $G$ to $K$ that agree on $H$ but not on all of $G$.

So far I've got nothing concrete. I did look at $N_G(H)$. I was thinking if two different element induce the same automorphism on $H$ i could maybe try inner automorphisms from them. And also other trivial cases like if the centralizer of $H$ has nontrivial elements etc. This has been my thought process so far.

I would appreciate any hints or help towards the solution.

Thank you

Ack
  • 379
  • Can $H$ be trivial? – Malady Jul 27 '24 at 20:17
  • 1
    Yes, but i believe we are already given an H – Ack Jul 27 '24 at 20:19
  • 2
    Do you know about the amalgamated free product construction? (AKA the pushout in the category of groups) – Alex Kruckman Jul 27 '24 at 20:20
  • Ah no, I am afraid not, if you take a look at the notes that I've linked then these are in the group actions section and I don't think the author meant that to be used – Ack Jul 27 '24 at 20:22
  • 2
    You can do this with $K=S_{|G|}$; think about the regular representation, and how you might reorder the points of $G$ so that $H$ still acts the same. – Steve D Jul 27 '24 at 20:47
  • I'm sorry but I don't seem to understand the part: "reorder points of G so that H acts the same", do we mean an automorphism of G? – Ack Jul 27 '24 at 20:52
  • Or I think $K = (G / H) \times (G / H)$ can work. – aschepler Jul 27 '24 at 21:27
  • 1
    IMO It's hard to give hints without giving away the full solution. By "reorder" I meant permuting $G$ (as a set!) via an inner automorphism of $S_G$. In order to ignore an annoying corner case, it's good to assume $H$ is non-normal in $G$; when $H\lhd G$ this is very easy. – Steve D Jul 27 '24 at 21:30
  • I just looked at the notes you linked; part (a) gives a version of what I've been hinting at. Have you done that part already? – Steve D Jul 27 '24 at 23:10
  • 1
    The existence of such a construction is equivalent to the fact that epimorphisms in the category of groups are surjective. – Geoffrey Trang Jul 28 '24 at 00:21
  • Yeah I had done part (a) I did consider cosets of H for part (b) but didn't see a way to make the group actions into a homomorphism – Ack Jul 28 '24 at 00:21
  • The construction here solves the problem. Though the claimed property is stronger than "every epimorphism is surjective", the proof establishes the stronger claim (that every subgroup is an equalizer subgroup). – Arturo Magidin Jul 28 '24 at 01:33

1 Answers1

3

I think part (a) of the exercise you linked gives an easy path to the solution.

Let $X$ be a set $G$ acts on transitively, and let $x_0\in X$, with $H\le G$ the stabilizer of $x_0$. If $\hat{X}\cong_G X$ but is disjoint from $X$, we can form $Y=X\sqcup\hat{X}$. There is a permutation $\pi$ of $Y$, swapping $x_0$ and $\hat{x}_0$, but fixing all other points. This permutation commutes with the action of $H$, but not of $G$.

Here's how to use it to solve part (b):

Let $X=G/H$ be the cosets of $H$ in $G$. As above, we get a homomorphism $\phi:G\rightarrow S_Y$. The element $\pi\in S_Y$ centralizes $\phi(H)$, but not $\phi(G)$. Thus if $\sigma$ is the inner automorphism that is conjugation by $\pi$, then $\sigma\phi:G\rightarrow S_Y$ is another homomorphism which agrees with $\phi$ on $H$, but not on $G$.

I had hinted at a similar way to solve this exercise in the comments. Let $\rho:G\rightarrow S_G$ be the regular representation. If $T$ is a transversal for $H$ in $G$, then any permutation $\tau\in S_T$ induces a permutation $\pi_\tau$ in $S_G$, by sending $th\mapsto \tau(t)h$. This permutation will fix each coset of $H$, and hence preserve the action of $H$ in the regular representation. To finish the problem, we need some $g\in G\setminus H$ that will act differently before/after applying $\pi_\tau$. This is always possible when $H$ is not normal in $G$, because then there exists $x\in G\setminus H$ and $y\in H$ such that $x^{-1}yx\not\in H$. Let $T$ be a transversal with $\{1,x\}\subset T$, and let $\tau=(1,x)$ be the transposition on $T$. Then for $h\in H$ and $tk\in TH$ we have \begin{equation} \pi_\tau\left(\rho(h)(tk)\right) = \pi_\tau(tkh) = \tau(t)kh = \rho(h)\left(\pi_\tau(tk)\right) \end{equation} However, if $yx=t_0h_0$, then \begin{equation} \pi_\tau\left(\rho(x)(y)\right) = \pi_\tau(yx)=\pi_\tau(t_0h_0)=\tau(t_0)h_0 \end{equation} Meanwhile, \begin{equation} \rho(x)\left(\pi_\tau(y)\right) = \rho(x)(xy) = x(yx) = xt_0h_0 \end{equation} For these to be equal, we must have $xt_0=\tau(t_0)$. Since $t_0\not\in\{1,x\}$, this is impossible.

Steve D
  • 5,737
  • After reading this I guess it does seem a bit more evident, although at the time I got the hint in the comments i actually had not much idea what to do with it. – Ack Jul 28 '24 at 00:30