The dihedral group $D_{16}$ of order 16 has a presentation $D_{16}= \langle a,t \ | \ a^2=t^8=atat=e\rangle$.

Question: Does there exist a finite 2-group $G$ containing $D_{16}$ as a subgroup, and an element $g \in G$ such that $gag^{-1}=t^4$? Bonus pats-on-the-back if $G$ has order 64.

An obvious reduction: one can assume that $G =\langle D_{16},g\rangle$.

An obvious constraint: $D_{16}$ cannot be normal in $G$ (so $G$ can't have order 32).

[This question has come up in investigations of the Balmer spectrum of $G$-equivariant stable homotopy for finite $p$-groups $G$. Like Dr. Frankenstein, I am looking for interesting subjects to experiment on, and my student Chris Lloyd is serving as the able assistant to the mad scientist.]

  • $\begingroup$ @ToddTrimble My comment in brackets in my original question was a link to algebraic topology. This is a problem about how finite p-groups, and their subgroups, interact with Morava K-theories, which are generalized homology theories. The existing literature focuses on abelian groups, but Lloyd and I now understand the problem for the dihedral group of order 8. This question, and a follow-up one that I posted after this one was so nicely answered, were related to understanding unusual aspects of a conjectural answer, as a help with choosing new non-abelian p-groups to study next. $\endgroup$ – Nicholas Kuhn Jul 4 '19 at 3:32
  • $\begingroup$ Thanks, Nicholas: you did indeed explain the connection; not sure why I missed that. I'll add the a-t tag back in, and I think you should feel free to rollback to the previous version, but maybe it doesn't matter too much after Derek's answer. $\endgroup$ – Todd Trimble Jul 4 '19 at 5:05

No. We can prove this by induction. Let $G$ be the smallest $2$-group in which this situation occurs. Then $G$ has a normal central subgroup $N$ of order $2$.

If $N$ has trivial intersection with the subgroup $\langle a,t \rangle = D_{16}$, then the same situation occurs in $G/N$, contradicting the minimality of $G$.

So that intersection must be nontrivial, and hence $N \le \langle a,t \rangle$, and then we must have $N = Z(\langle a,t \rangle) = \langle t^4 \rangle$.

But then $t^4 \in Z(G)$, contradicting the assumption that it is conjugate in $G$ to $a$.

The situation you describe can occur in finite groups, such as in simple groups ${\rm PSL}(2,q)$ for prime powers $q$ with $q \equiv 15$ or $17 \bmod 32$, ($q=17$ for example), but not in finite $2$-groups.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks for the friendly proof! I really have a more general question of the form "Can this happen?" and this question here was my attempt to find an example. If I can't easily resolve my general question with arguments in the style of what you did, I'll be back with more! $\endgroup$ – Nicholas Kuhn Jun 27 '19 at 23:44
  • 1
    $\begingroup$ Also thanks for the composite order group example! $\endgroup$ – Nicholas Kuhn Jun 28 '19 at 0:38
  • 4
    $\begingroup$ The question, and Derek's elegant answer and example me wonder whether it is possible to give a reasonable answer to "Which 2-groups S can be embedded in a finite group G with one conjugacy class of involutions?" More ambitiously, one might try to insist that $G$ be solvable. I think these are different questions ( or give different answers anyway): I think a semidihedral $2$-group of order 16 embeds in $M_{11}$ which has one class of involutions, but not in any finite solvable group with one class of involutions. $\endgroup$ – Geoff Robinson Jun 28 '19 at 11:22
  • 3
    $\begingroup$ @Geoff: If I recall correctly, Thompson has proved that a solvable group with one class of involutions has $2$-length one. $\endgroup$ – Richard Lyons Jun 28 '19 at 12:40
  • 3
    $\begingroup$ Without putting any constraint on the finite group, problems like this can always be solved. If $G$ is any finite group, and $G\rightarrow S$ is the embedding of $G$ into the group of all permutations of the set $G$, then any isomorphism between subgroups of $G$ is a conjugation inside $S$. I'm not sure whether $S_{16}$ is simpler than Derek's $PSL(2,15)$ as a non 2-group solution to the original question. This doesn't help with Geoff's question. $\endgroup$ – IJL Jul 2 '19 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.