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In this recent question, Iota asked if, given a finite group $G$ and two isomorphic normal subgroups $H$ and $K$, it would follow that $G/H$ and $G/K$ are isomorphic. This is not true (a simple example given by $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, with $H=\langle (1,0)\rangle$ and $K=\langle (0,2)\rangle$). A sufficient condition to guarantee isomorphic quotients is the existence of an automorphism $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H)=K$.

One can also have $G/H\cong G/K$, but $H$ and $K$ not isomorphic. For example, take again $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, and take $H=\mathbb{Z}_2\oplus\langle (0,2)\rangle$, isomorphic to the Klein $4$-group, and $K=\{0\}\oplus\mathbb{Z}_4$. Then $G/H\cong G/K\cong \mathbb{Z}_2$.

And of course, it is trivial to have $H\cong K$ and $G/H\cong G/K$.

Now, here's the question:

Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$?

It is not hard to come up with examples with infinite $G$. For example, take $$G= \mathbb{Z}_2\oplus \left(\bigoplus_{i=1}^{\infty}\;\mathbb{Z}_4\right),$$ let $H=2G$, and let $K$ be the subgroup generated by $H$ and the generator of the cyclic factor $\mathbb{Z}_2$. Then both $H$ and $K$ are isomorphic to a direct sum of countably many copies of $\mathbb{Z}_2$, as are the quotients $G/H$ and $G/K$. But $H$ is a verbal subgroup, hence fully invariant, so any automorphism of $G$ maps $H$ to $H$, so there does not exist any $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H)=K$.

Does someone have an example with finite $G$? Note that I am not requiring that $\varphi$ be a lift of the given isomorphism of $G/H$ with $G/K$.

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    $\begingroup$ I think there could be a canonical isomorphism between $H$ and $K$ in this case ; take the isomorphism between $H$ and $K$, name it $\varphi_1$, and in the same manner, name the isomorphism between $G/H$ and $G/K$ $\varphi_2$. Now take an arbitrary element of $G$, look in which class of $G/H$ it lies in and write $g = g_h h$, and map $h$ by $\varphi_1$, $g_h$ by $\varphi_2$. I think there's some way you can construct an automorphism in this manner. $\endgroup$ May 23, 2011 at 18:30
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    $\begingroup$ @Patrick: That will not work in general. Take $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, take $H=K$ generated by $(0,2)$, let $\varphi_1$ be the identity, and let $\varphi_2$ be the map that swaps the two coordinates of $G/H\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$. If we take $g=(0,1)$, written as $(0,1)+(0,0)$, then the image of $(0,1)+H$ is $(1,0)+K$, but we cannot map $(0,1)$ to either element of that coset (neither $(1,0)$ nor $(1,2)$), because both are of order $2$, whereas $(0,1)$ is of order $4$. Of course, here $\varphi$ exists, but the point is your construction does not. $\endgroup$ May 23, 2011 at 18:46
  • $\begingroup$ Good to know. I was actually trying to work it out and got stuck at the part where I had to prove my thing was a bijection so... good to know I won't be stuck any longer. ^^ $\endgroup$ May 23, 2011 at 18:48
  • $\begingroup$ Here is a similar question with modules: math.stackexchange.com/questions/89732/… $\endgroup$ Jan 26, 2012 at 22:05
  • $\begingroup$ @ArturoMagidin Can the question in your post be thought about using HNN extensions? $\endgroup$
    – user5826
    Nov 17, 2018 at 2:07

1 Answer 1

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There are lots of examples, see here:

http://groupprops.subwiki.org/wiki/Series-equivalent_not_implies_automorphic

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  • $\begingroup$ I was about to copy paste your email over here. Anyway welcome to the site. Hope you have a nice time. $\endgroup$
    – user9413
    May 23, 2011 at 18:46
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    $\begingroup$ Aha; in particular, the case of the abelian group $\mathbb{Z}_{p^3}\oplus\mathbb{Z}_{p^2}\oplus\mathbb{Z}_p\oplus\mathbb{Z}_p$ should have suggested itself. Clearly, I needed to think more about this. $\endgroup$ May 23, 2011 at 18:51
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    $\begingroup$ I don't think the examples are obvious; the abelian example wouldn't have struck me offhand. $\endgroup$
    – Vipul
    May 23, 2011 at 19:45

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