Automorphism group of the quaternion group

Question

Let $Q_8$ be the quaternion group. How do we determine the automorphism group ${\rm Aut}(Q_8)$ of $Q_8$ algebraically? I searched for this problem on internet.

I found some geometric proofs that ${\rm Aut}(Q_8)$ is isomorphic to the rotation group of a cube, hence it is isomorphic to the symmetric group $S_4$.

I would like to know an algebraic proof that ${\rm Aut}(Q_8)$ is isomorphic to $S_4$.

Answer 1

$Q_8$ has three cyclic subgroups of order $4$: $\langle i\rangle$, $\langle j\rangle$, $\langle k\rangle$, and ${\rm Aut}(Q_8)$ acts on these three subgroups; inducing a homomorphism $\Phi\colon{\rm Aut}(Q_8)\rightarrow S_3$. We can see that, the homomorphism is surjective, since the two automorphisms $f\colon i\mapsto j, j\mapsto i$, and $g\colon j\mapsto k, k\mapsto j$ give two transpositions in $S_3$.

The kernel contains those $\varphi\in{\rm Aut}(G)$ such that $\varphi(\langle i\rangle)=\langle i\rangle$ and $\varphi(\langle j\rangle)=\langle j\rangle$ (automatically, $\varphi(\langle k\rangle)=\langle k\rangle$).

(1) $\varphi(\langle i\rangle)=\langle i\rangle$ means $\varphi(i)\in \{i,-i\}$, and similarly, $\varphi(j)\in \{j,-j\}$. One can check that these four choices are automorphisms of order $2$ (or $1$) (since they are switching elements in a pair), and hence kernel is Klein-$4$ group $V_4$.

(2) Since, automorphisms in the kernel fix all the cyclic subgroups of $Q_8$ (not necessarily point-wise); consider the following automorphisms:

$S: i\mapsto j, j\mapsto i$, (hence $k\mapsto -k$) and

$T\colon j\mapsto k, k\mapsto j$ (hence $i\mapsto -i$);

these are not inner (since they fix a subgroup), and they generate $S_3$ (they are like transpositions $(1\,2)$ and $(2\,3)$ ). Therefore, we have $\langle S,T \rangle=K\leq{\rm Aut}(Q_8)$, such that $K\cong S_3$ and $\Phi(K)=S_3$. Also,

$\ker(\Phi) \cap K=\phi $.

Therefore, ${\rm Aut}(Q_8)=\ker(\Phi)\rtimes K \cong V_4\rtimes S_3$.

Consider an element of $\ker(\Phi)$:

$f:i\mapsto -i$, $j\mapsto j$,

and two elements of $K\cong Im$:

$g\colon i\mapsto j, j\mapsto i $ (like a transposition), and $h\colon i\mapsto j, j\mapsto k$ (like a $3$-cycle).

One can check that $f$ doesn't commute with $g$ as well as $h$.

In fact, this shows that no element of $V_4\setminus\{1\}$ commutes with any element of $K\setminus \{ 1\}$ (by inter-changing roles of $i,j,k$); this means, the action of $K$ on $V_4$ (by conjugation) is faithful; and up to equivalence, there is only one such action. Therefore, ${\rm Aut}(Q_8)=V_4\rtimes K\cong S_4$

Answer 2

Hint:

$Inn(Q_8)\cong V$ and it equals its own centralizer in $Aut(Q_8)$. Now use $N/C$ Lemma in which $G=Aut(Q_8)$ and $H=Inn(Q_8)$. Of course using the lemma we always have $Inn(G)\vartriangleleft Aut(G)$ and $G/Z(G)\cong Inn(G)$ in which $G$ is our group.

Answer 3

You can find the proof here: automorphism of generalized quaternionic group

See Proposition 1.1 (pg. 156). Your case can be obtained taking $m=2$. But they prove for $m\geq 2$.

Answer 4

OK, let's first put an upper bound on the number of automorphisms of $Q_8$.

There are $6$ elements of order $4$ in $Q_8$. It is well-known that $i$ and $j$ generate $Q_8$, so their images under some automorphism $\phi$ are enough to determine that automorphism uniquely. So there are $6$ choices for $\phi(i)$. Now we cannot have $\phi(j)\in\langle\phi(i)\rangle$, because then $\phi$ would fail to be surjective. Thus there are $6-2=4$ possibilities for $\phi(j)$. This crude reasoning gives the upper bound of $6\cdot4=24$ automorphisms.

Let $\alpha$ be any permutation on the elements $\lbrace i,j,k\rbrace$ (as a set). We can "extend" $\alpha$ to a map of $Q_8$ in the obvious way (let it commute with negatives). Since any two of $\lbrace i,j,k\rbrace$ generates $Q_8$, this map will be surjective, and since $Q_8$ is finite, injective as well. It only remains to show it is a homomorphism. Since $i^2=j^2=k^2=-1$ is central and the only element of order $2$, we only need show $$ \alpha(i)\alpha(j)=\alpha(k);$$

This is not always true, but what is always true is that $$ \alpha(i)\alpha(j)=\pm\alpha(k),$$ and so we see each permutation gives rise to an automorphism (after a suitable choice of sign).

Since the subgroups $\langle i\rangle$,$\langle j\rangle$,$\langle k\rangle$ are normal, none of these permutations is an inner automorphism. Thus they give a subgroup isomorphic to $S_3$ inside $Aut(Q_8)$, which trivially intersects $Inn(Q_8)\cong Q_8/Z(Q_8)\cong C_2\times C_2$. Thus $Aut(Q_8)$ has size at least $|S_3|\cdot |C_2\times C_2|=24$ automorphisms.

Since $Inn(Q_8)\lhd Aut(Q_8)$, we get a semidirect product $(C_2\times C_2)\rtimes S_3$ inside $Aut(Q_8)$, which is then the whole group $Aut(Q_8)$. This is easily seen to be isomorphic to $S_4$, and so we are done.