A finite 2-group containing the dihedral group of order 16?

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.]

• @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. – Nicholas Kuhn Jul 4 '19 at 3:32
• 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. – 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.

• 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! – Nicholas Kuhn Jun 27 '19 at 23:44
• Also thanks for the composite order group example! – Nicholas Kuhn Jun 28 '19 at 0:38
• 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. – Geoff Robinson Jun 28 '19 at 11:22
• @Geoff: If I recall correctly, Thompson has proved that a solvable group with one class of involutions has $2$-length one. – Richard Lyons Jun 28 '19 at 12:40
• 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. – IJL Jul 2 '19 at 9:12