Show ${\rm Aut}(G)$ is a $2$-group, where $G$ is given by a particular presentation

This is Exercise 5.3.5 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

The exercise is marked as being referred to later on in the text.

NB: I have not used the combinatorial-group-theory tag for a reason. The tools available for presentations in the book so far are very limited.

The Details:

For a given prime $p$, a $p$-group is a group such that the order of every element is a power of $p$.

A group presentation $\langle X\mid R\rangle $ is the quotient of the free group $F_X$ generated by the elements of $X$ by the normal subgroup $\langle\langle R\rangle\rangle$ of $F_X$ generated by $R$, where $R$ is a set of words in $X\cup X^{-1}$; such words can be written as $u=v$ for $uv^{-1}$.

The automorphism group ${\rm Aut}(G)$ of a group $G$ is the group of all isomorphisms from $G$ to $G$ under function composition.

The Question:

Paraphrased:

Prove that ${\rm Aut}(G)$ is a $2$-group, where

$$G\cong\langle x,y\mid x^2, y^{2^n}, y^x=y^{1+2^{n-1}}\rangle.$$

Here $y^x$ is defined to be $x^{-1}yx$.

Thoughts:

If we kill $y^2$ (that is, take the quotient of $G$ by $\langle\langle y^2\rangle\rangle$), we get

$$\begin{align} G/\langle\langle y^2\rangle\rangle &\cong \langle x,y\mid x^2, y^2, yx=xy\rangle\\ &\cong \Bbb Z_2\times\Bbb Z_2. \end{align}$$

That's a $2$-group. This doesn't get me anywhere though, especially since ${\rm Aut}(\Bbb Z_2\times\Bbb Z_2)\cong S_3$, which is not a $2$-group.

I have experimented in GAP. Here is some of the code I used:

RobinsonExerciseGroup:=function(n)
    local F, rels;
    F:=FreeGroup(2);
    rels:=[(F.1)^2, (F.2)^(2^n), (F.1)^(-1)*(F.2)*(F.1)*(F.2)^(-1-(2^(n-1)))];
    return F/rels;
end;

I checked the order of ${\rm Aut}(G)$ up to $n=8$, where the calculation got too much for my laptop. They're all powers of two. I checked the structure of the automorphism group up to $n=8$, too, but I didn't spot any pattern.

I know that each automorphism $\varphi:G\to G$ is completely determined by its behaviour on $x$ and $y$; that is, if we know $\varphi(x)$ and $\varphi(y)$, then we know $\varphi(g)$ for all $g\in G$.

Please help :)

Solution 1:

You seem to be on the right track, so I will make this a sequence of hints rather than a detailed solution.

Let $A = {\rm Aut}(G)$. In fact $Z := \langle y^2 \rangle = Z(G)$, so $A$ must fix the subgroup $Z$, and it makes sense to look at the action of $A$ on $G/Z \cong C_2 \times C_2$, as you are doing.

Now $A$ has a normal subgroup $A_I$ consisting of those automorphisms that induce the identity on $G/Z$, and $A/A_I$ is isomorphic to the induced action of $A$ on $G/Z$. We need to prove that both $A_I$ and $A/A_I$ are $2$-groups.

$A_I$ consists of the so-called central automorphisms. Why don't you have a go at proving that $A_I$ is a $2$-group?

As for $A/A_I$, as you remarked, ${\rm Aut}(G/Z) \cong S_3$. However, an automorphism of order $3$ would induce a $3$-cycle of the cosets $xZ$, $yZ$ and $xyZ$. Try and prove that no automorphism of $G$ can map $x$ to an element of $yZ$.