# Showing ${\rm Aut}(Q_{2^n})\cong{\rm Hol}(\Bbb Z_{2^{n-1}})$ for $n>3$

This is part of Exercise 5.3.4 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.

The second part is here:

For previous questions of mine on generalised quaternion groups, see here:

## The Details:

On page 37 of the book,

Let $$\lambda:G\to {\rm Sym}\, G$$ and $$\rho:G\to{\rm Sym}\, G$$ be the left and right regular representations of a group $$G$$. Then $$G^\lambda$$ and $$G^\rho$$ are subgroups of $${\rm Sym}\, G$$, as is $${\rm Aut}(G)$$. Now $$g^\lambda g^\rho$$ maps $$x$$ to $$g^{-1}xg$$, so $$g^\lambda g^\rho$$ is just $$g^\tau$$, the inner automorphism induced by $$g$$. Consequently

$$\langle G^\lambda, {\rm Aut}(G)\rangle =\langle G^\rho, {\rm Aut}(G)\rangle;$$

this subgroup of $${\rm Sym}\, G$$ is known as the holomorph of the group $$G$$,

$${\rm Hol}\, G.$$

Elsewhere in the book, we have:

$$D_{2^n}\cong \langle r,s\mid r^{2^{n-1}}, s^2, srs=r^{-1}\rangle$$

is the dihedral group of order $$2^n$$ and

$$Q_{2^n}\cong\langle x,y\mid x^{2^{n-1}}, y^2=x^{2^{n-2}}, y^{-1}xy=x^{-1}\rangle$$

is the generalised quaternion group of order $$2^n$$ (defined for $$n\ge 3$$).

## The Question:

Prove that $${\rm Aut}(Q_{2^n})\cong {\rm Hol}(\Bbb Z_{2^{n-1}})$$ if $$n>3$$.

## Thoughts:

$${\rm Aut}(D_{2^n})\cong{\rm Aut}(Q_{2^n})$$

for $$n>3$$. It might help.

By definition, we have

$${\rm Hol}(\Bbb Z_{2^{n-1}})=\langle \Bbb Z_{2^{n-1}}^\lambda, {\rm Aut}(\Bbb Z_{2^{n-1}})\rangle.$$

Here $$\Bbb Z_{2^{n-1}}^\lambda$$ is the image of $$\Bbb Z_{2^{n-1}}$$ under $$\lambda$$, i.e.,

$$\Bbb Z_{2^{n-1}}^\lambda=\{ g^\lambda: x\mapsto g^{-1}x\mid g\in \Bbb Z_{2^{n-1}}\}$$

and it is well-known that

$${\rm Aut}(\Bbb Z_{2^{n-1}})\cong U(2^{n-1}),$$

where $$U(m)$$ is the group of units modulo $$m$$. (See Theorem 6.5 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".)

That's all I have there.

In GroupNames, we have

• $${\rm Aut}(Q_{2^4})\cong \langle a,b,c\mid a^8=b^2=c^2=1, bab=a^3, cac=a^5, bc=cb\rangle\cong \Bbb Z_8\rtimes\Bbb Z_2^2$$.

• $${\rm Aut}(Q_{2^5})\cong\langle a,b,c\mid a^{16}=b^2=c^4=1, bab=a^{-1}, cac^{-1}=a^5, bc=cb\rangle\cong D_{32}\rtimes \Bbb Z_4$$.

• $$\lvert{\rm Aut}(Q_{2^6})\rvert=512$$.

• $$\lvert{\rm Aut}(Q_{2^7})\rvert=2,048$$.

• The previous question basically describes all the automorphisms and gives you the solution to this question. Look at e.g. the automorphisms $\alpha(a,b)$ in the answer by Derek Holt. Calculate $\alpha(a,b) \circ \alpha(a',b')$. This is equal to $\alpha(a'',b'')$ for some $a''$ and $b''$, what are $a''$ and $b''$?
• Yes I agree, it follows easily from my previous answer. The elements of the automorphism group with $a=1$ form a normal subgroup that is isomorphic to $C_{2^{n-1}}$, and the elements with $b=0$ form a complementary subgroup isomorphic to ${\rm Aut}(C_{2^{n-1}})$. Commented Jan 8, 2022 at 9:23