# Abelian subgroup contained in a normal subgroup; a trick about elements of order $3$

Let $A$ be a normal subgroup of $G$. Suppose that every element in $G\,\backslash\, A$ has order $\bf 3$. Then $[B,B^x]=1$ for all Abelian subgroups $B\leq A$ and $x\in G\,\backslash\, A$.

I have been told that my task must have something to do with Chapter VI, On the isomorphism of a Group with Itself, para 66. of the famous book—“Burnside, W.: Theory of Groups of Finite Order, 2nd edn., Cambridge 1911; Dover Publications, New York 1955”. [ It’s a trick about order $3$, which was mentioned in the comments and Derek Holt’s answer below. ]

Although we‘ve made many attempts indeed and have made a breakthrough (Derek Holt’s answer), yet we haven’t been able to figure out how to use the normality of $A$ and abelianity of $B$, on which I’m still struggling...

It would be greatly appreciated if you are kind enough to provide a reasonable answer!

PS: It’s exercise 1.5.6 of the book The Theory of Finite Groups, An Introduction. Berlin: Springer, 2004.

• Page 90 of Burnside contains his celebrated treatment of finite groups with fixed-point-free automorphisms of order 2 (they are abelian) or of order 3 (they are direct products of their Sylow subgroups, he references his paper Proc LMS XXXV (1902) pp 28 - 37). It is a very long time since I struggled with this page. I will add what I think is the crucial argument in a moment. – ancientmathematician Mar 2 '18 at 11:35
• This, if I recall, is the crucial part of Burnside's argument. If $a\in A$ and $x\not\in A$ then $a a^x a^{x^2}=(a x^2)^3=1$ and so $a a^{x^2} a^x=a a^{x^2} a^{x^4}=1$, so that $a^x$ and $a^{x^2}$ commute; similarly $a$ and $a^x$ commute. – ancientmathematician Mar 2 '18 at 11:40
• It has to be the 2nd edition. It is Chapter VI, On the Isomorphism of a group with itself, It is para 66. – ancientmathematician Mar 2 '18 at 11:52
• I think by this he just means the permutation carries the group element $S$ [traditionally, a "substitution"] to the group element $S'$. – ancientmathematician Mar 2 '18 at 12:18
• I think that the Burnside trick (as you call it) lets one prove that $\langle B,B^x\rangle$ is fixed by $x$. Moreover I think it lets one prove $BB^{x}=BB^{x^2}$(*); and then $B^x B=B^{x}(B^{x})^{x^2}=B^x B^{x^2}=B B^{x^2} B^{x^2}=B B^{x^2}=B B^{x}$ by repeated use of (*) and the trick. Now without loss of generality one can assume $A=B B^x$, $G=A\langle x\rangle$. I note that one is then in the same situation as the previous exercise, the Ito one. I am now stuck. – ancientmathematician Mar 3 '18 at 8:22

## 1 Answer

This is not a complete answer, but I hope it helps.

Let $b \in A$ and $x \in G \setminus A$. We claim that $b$ commutes with $b^x = x^{-1}bx$.

Let $u = bx^{-1}bx$ and $v = x^{-1}bxb$. We want to show that $u=v$.

Now, since $b\in A,\,x\not\in A\Rightarrow bx^{-1}\not\in A\Rightarrow(bx^{-1})^3=1$, we have $(bx^{-1})^2u = bx$.

Similarly, using $x^{-2}=x$ and $(bx)^3=1$, we get $(bx^{-1})^2v = bx^{-1}bxbxb = bx^{-1}(bx)^3x^{-1} = bx^{-2} = bx$.

So $(bx^{-1})^2u = (bx^{-1})^2v$ and hence $u=v$, as claimed.

Added March 20: As Zach Teitler has pointed out, I had answered this question here. So I am afraid it is a duplicate.

• Isn't this just Burnside's argument, @Derek ? – ancientmathematician Mar 2 '18 at 13:17
• Yes it is. But we need to prove that $a$ commutes with $b^x$ for any $a,b \in A$ with $ab=ba$ and $x \in G \setminus A$, and I cannot think how to do that, although I suspect that must be some slick proof! – Derek Holt Mar 2 '18 at 13:44
• But now that we can see where the problem comes in K & Stellmacher it must be much easier than I thought. – ancientmathematician Mar 2 '18 at 17:11
• @DerekHolt Perhaps you have forgotten that you answered this in 2015: math.stackexchange.com/questions/1484993/… :-) – Zach Teitler Mar 20 '18 at 18:18
• @ZachTeitler Yes I had forgotten, thanks! So the question is actually a duplicate. – Derek Holt Mar 20 '18 at 18:46