# Forced conjugation of elements in finite groups

The dihedral group $$D_8 = \langle\ a,\ b \ \mid \ a^4,\ b^2,\ (ab)^2\ \rangle$$ has a central involution $c=a^2$ and a non-central one, $b$.

Q. Can we embed $D_8$ into a finite group $G$ in which $c$ and $b$ become conjugate?

• To make the question less basic, one cas ask whether one can get $G$ to be a 2-group. – YCor Jun 16 '17 at 11:09
• @YCor The answer seems to no. Prove it by induction on $|G|$. Let $N$ be a central subgroup of $G$ of order $2$. If $N \le D_8$ then the answer is clearly no. Otherwise apply induction to the image of $D_8$ in $G/N$. – Derek Holt Jun 16 '17 at 11:55
• @DerekHolt thanks for the nice simple argument! – YCor Jun 16 '17 at 13:05

Actually for any finite group $H$, any two elements of the same order are conjugate in some larger finite group $G$.

Proof: Consider the embedding into the symmetric group given by the permutation action on elements of $H$ by left multiplication. An element of order $n$ has $|H|/n$ orbits, all of size $n$, which uniquely determines the conjugacy class.

• Very nice! Another, less explicit but perhaps more conceptual, proof, is to notice that the universal group in which $b$ and $c$ are conjugate, namely the HNN extension of $H$ conjugating $b$ and $c$, is virtually free and hence residually finite. – HJRW Jun 16 '17 at 8:29
• I would also have thought of the same argument as Henry's, which is possibly useful if one wants to have a reasonable upper bound on the order of the finite group in which we get an embedding. – YCor Jun 16 '17 at 10:31
• @YCor, could you say a bit more about the upper bound that this argument yields? – verret Jun 16 '17 at 10:44
• The bound coming from Will Sawin's argument (originally due to B.H. Neumann) is $|H|!$. It might be difficult to improve on that in general. – Derek Holt Jun 16 '17 at 11:09
• @verret I'm just saying that if one wishes to get the best bound, this sounds a natural approach: one has to study finite index subgroups of the given HNN extension, which, in the current case, should be small enough to be amenable to computer computations. In a sense it's the universal approach since any group generated by $H$ and the given conjugating element is quotient of this HNN extension. – YCor Jun 16 '17 at 11:14

Regarding your specific example, the dihedral group of order $8$ admits an embedding into $S_4$, namely $$D_8 \simeq \langle (1 \, 2 \, 3 \, 4), \, (1 \, 3) \rangle.$$ The center is generated by $(1 \, 3)(2 \, 4)$, and the remaining involutions are $(2 \, 4)$, $(1\, 2)(3 \,4)$, $(1 \, 4)(2 \, 3)$, $(1 \, 3)$.

Then in $S_4$ the central involution of $D_4$ becomes conjugate to two non-central ones.

• Note that there is an outer automorphism that swaps the two conjugacy classes of noncentral involutions, so this method handles both. – Will Sawin Jun 16 '17 at 12:09

Will Sawin's lovely answer deals with the question in much greater generality, but for the specific case in hand, many Finite Group Theory texts ( eg Gorenstein (1968)) use groups with dihedral Sylow $2$-subgroups to illustrate possible fusion patterns via Alperin's fusion theorem. I will discuss the possibilities in the case of a finite group $G$ with a dihedral Sylow $2$-subgroup $D$ of order $8$ (ie with $8$ elements). There are three different possibilities. Note that $D$ has two different Klein $4$-subgroups $U$ and $V$, and that since ${\rm Aut}(D)$ is a $2$-group, we have $G = DC_{G}(D).$ By a Theorem of Burnside, $U$ and $V$ are not conjugate in $G$ (since they are certainly not conjugate in $N_{G}(D)$ and both are normal in $D).$ The three possibilities are:

1. $N_{G}(U)/C_{G}(U) \cong N_{G}(V)/C_{G}(V) \cong \mathbb{Z}/2\mathbb{Z}.$ In this case, $G$ has a normal $2$-complement, and involutions of $D$ are conjugate in $G$ if and only if they are already conjugate in $D.$

2. $N_{G}(U)/C_{G}(U) \cong \mathbb{Z}/2\mathbb{Z}$ and $N_{G}(V)/C_{G}(V) \cong S_{3}.$ In this case, $G$ does not have a normal $2$-complement but does have a normal subgroup of index $2.$ In this case, all the involutions of $V$ all become conjugate in $G,$ but not all involutions of $U$ are $G$-conjugate. The example of $G = S_{4}$ given by Francesco Polizzi gives an example where this occurs.

3. $N_{G}(U)/C_{G}(U) \cong N_{G}(V)/C_{G}(V) \cong S_{3}.$ In this case, $G$ has no factor group of order $2$ and all $5$ involutions of $D$ are conjugate in $G$. The examples of $G \cong A_{6}$ or $G \cong {\rm PSL}(2,7)$ are cases where this occurs (there are many more examples, of course).

Later edit: regarding the exchange in comments between @YCor and @Derek Holt : it is clear from this analysis that $S_{4}$ is the smallest group with a subgroup $D_{8}$ such that a non-central involution of $D_{8}$ and a central involution of $D_{8}$ becoming conjugate in the overgroup. Also ${\rm PSL}(2,7)$ is the smallest group with a subgroup $D_{8}$ whose involutions are all conjugate in the overgroup.

(1) (essentially HJRW's comment to Will Sawin's answer)

Given a finite group $F$ and an isomorphism $t$ between two subgroups $A,B$ of $F$, the universal group in which $A$ and $B$ are conjugate through $t$, namely the HNN extension $H$ of $(F,A,B,t)$, is virtually free and hence residually finite. Thus, there exists a finite quotient of $H$ in which $F$ is mapped injectively, and in this quotient $A$ and $B$ are indeed conjugate (by $t$). This applies in particular to the case when $A,B$ are cyclic subgroups of the same order.

Of course this argument, which looks somewhat immediate at first glance, is less elementary than the one given by Will Sawin since it relies on residual finiteness of HNN extensions of finite groups.

(2) Will's argument extends to this setting ($F$, $A$, $B$ without assuming $A,B$ cyclic. Namely, $A$ has two free actions on $F$, one being given by $a\cdot g=ag$, the other by $a\cdot g=t(a)g$. Writing $k=|F/A|$, find $k$ points $x_1,\dots,x_k$, one in each orbit of the first action, and $k$ points $y_1,\dots,y_k$, one in each orbit of the second action. Then extend the assignment $x_i\mapsto y_i$ to a permutation $\sigma$ of $F$, by the assignment $\sigma(ax_i)=t(a)y_i$. This is well-defined by freeness of the action. Then $$\sigma(abx_i)=t(ab)y_i=t(a)t(b)y_i=t(a)\sigma(bx_i)$$ for all $a,b\in A$ and $i$, and thus $\sigma(ag)=t(a)\sigma(g)$ for all $a\in A$ and $g\in F$. In other words, $\sigma\circ L_a=L_{t(a)}\circ \sigma$. So in the permutation group of $F$, where $F$ is identified to its image through left multiplication, the isomorphism $t:A\to B$ is realized by conjugation by $\sigma$.