Automorphism group of the quaternion group 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 $S_4$.
 A: 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.
A: 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.
A: Here is just an elaboration of Beginner's and user641's answers:
Using the presentation
$$Q_8=\langle a,b\mid a^4=1, \ a^2=b^2, \ b^{-1}ab=a^{-1}\rangle,$$
one can show the following:
Proposition 1. Let
$$A\mathrel{\mathop:}=\{(a,b)\in Q_8^2\mid |a|=4,\ |b|=4,\ a\neq b,\ a\neq b^{-1}\}.$$
Then there is a bijection
$$\Psi:\operatorname{Aut}(Q_8)\to A,\qquad\xi\mapsto(\xi(i),\xi(j)).$$
Using this proposition, one can actually construct the automorphisms on $Q_8$. In particular, counting the number of elements of $A$ gives $|\operatorname{Aut}(Q_8)|=24$.
$Q_8$ has exactly 3 subgroups of order 4, namely $\langle i\rangle$, $\langle j\rangle$, and $\langle k\rangle$. Hence, $\operatorname{Aut}(Q_8)$ acts on $A\mathrel{\mathop:}=\{ \langle i\rangle, \langle j\rangle, \langle k\rangle\}$. Let
$$\pi:\operatorname{Aut}(Q_8)\to S_A$$
be the associated permutation representation. For brevity, let us write $1\mathrel{\mathop:}=\langle i\rangle$, $2\mathrel{\mathop:}=\langle j\rangle$, $3\mathrel{\mathop:}=\langle k\rangle$, so that $A=\{1,2,3\}$ and $S_A=S_3$.
Now, there is an isomorphism
$$S_3\xrightarrow{\sim}\langle a,b\mid a^2=b^2=1,\ (ab)^3=1\rangle,$$
such that $(12)\mapsto a$ and $(13)\mapsto b$.
Using this presentation, one can show that there is a group homomorphism
$$\Psi:S_3\to\operatorname{Aut}(Q_8)$$
such that $(12)\mapsto\tau$ and $(23)\mapsto\sigma$, where
$$\tau\mathrel{\mathop:}=(ij)(-i-j)(k-k)\qquad\mbox{and}\qquad\sigma\mathrel{\mathop:}=(i-i)(jk)(-j-k),$$
if you will pardon my cycle notation. (It doesn't look nice, but there is no ambiguity.)
One can verify that
$\pi\circ\Psi$ is the identity map on $S_3$. This implies that $\pi$ is surjective and $\Psi$ is injective. Also, $\operatorname{Im}\Psi\cap\ker\pi=1$.
$\pi$ therefore factors through the isomorphism
$$\frac{\operatorname{Aut}(Q_8)}{\ker\pi}\xrightarrow{\sim}S_3,$$
from which it follows that $|\ker\pi|=4$.
Indeed, using Propsition 1, one can see that
$$\ker\pi=\{1,\ (i-i)(k-k),\ (j-j)(k-k),\ (i-i)(j-j)\}\simeq V_4,$$
the Klein four-group.
Now, every subgroup of $Q_8$ is normal, so
$$\operatorname{Inn}(Q_8)\subseteq\ker\pi.$$
On the other hand, $Z(Q_8)=\{1,-1\}$ and
$$\operatorname{Inn}(Q_8)\simeq\frac{Q_8}{Z(Q_8)},$$
so $|\operatorname{Inn}(Q_8)|=4$. This implies that
$$\operatorname{Inn}(Q_8)=\ker\pi.$$
Let $H\mathrel{\mathop:}=\operatorname{Im}\Psi$. Since $H\cap\operatorname{Inn}(Q_8)=1$,
$$|\operatorname{Inn}(Q_8)H|=|V_4||S_3|=4\cdot6=24=|\operatorname{Aut}(Q_8)|,$$
i.e.,
$$\operatorname{Inn}(Q_8)H=\operatorname{Aut}(Q_8).$$
Using Proposition 1 once again, one can see that
$$\eta\mathrel{\mathop:}=(i-j)(-ij)(k-k)$$
is an automorphism on $Q_8$.
There is an isomorphism
$$S_4\xrightarrow{\sim}\langle a,b,c\mid a^2=b^2=1, \ (ab)^3=(bc)^3=1, \ (ac)^2=1\rangle$$
such that $(12)\mapsto a$, $(23)\mapsto b$, and $(34)\mapsto c$.
Using this presentation, one can show that there is a group homomorphism
$$\Phi:S_4\to\operatorname{Aut}(Q_8)$$
such that $(12)\mapsto\tau$, $(23)\mapsto\sigma$, and $(34)\mapsto\eta$.
Note that $\Phi|_{S_3}=\Psi$, so $\Phi$ restricts to an isomorphism
$$S_3\xrightarrow{\sim}H.$$
Also, one can check that $\Phi$ restricts to an isomorphism
$$V\xrightarrow{\sim}\operatorname{Inn}(Q_8),$$
where
$$V\mathrel{\mathop:}=\{1,(12)(34),(13)(24),(14)(23)\}\subseteq S_4$$
is a subgroup.
(Counting the number of elements in the conjugacy classes of $S_4$ reveals that the only nontrivial proper normal subgroups of $S_4$ are $V$ and $A_4$.)
Note that $V\cap S_3=1$, so $S_4=VS_3$.
Also, $\operatorname{Aut}(Q_8)=\operatorname{Inn}(Q_8)H$, so the restricted isomorphisms $V\simeq \operatorname{Inn}(Q_8)$ and $S_3\simeq H$ imply that $\Phi$ is surjective, i.e., it is an isomorphism.
A: $Q_8$ 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, ${\rm Aut}(Q_8)=V_4\rtimes K\cong S_4$
A: The simplest way to determine the automorphism group of $Q_{8}$ may be to realize that $$Q_{8} \ \text{char} \ SL(2,3) \unlhd GL(2,3)$$ So $Q_{8} \unlhd GL(2,3)$. One can take $$\begin{equation*} x =\left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right), \ \ y = \left( \begin{array}{cc} 1 & -1 \\ -1 & -1 \end{array} \right) \end{equation*}$$ as the two generators of $Q_{8}$. 
Let $G=GL(2,3)$ and $Q=Q_{8}$. Then $C_{G}(Q) = Z(G)$ and therefore $PGL(2,3)=G/Z(G)$ is isomorphic to a subgroup of $Aut(Q)$. Now $|G|=48$ and $|Z(G)|=2$. Thus $|G/Z(G)|=24$. As others have shown $|Aut(Q)| \leq 24$. Hence we can conclude that $$PGL(2,3) \simeq Aut(Q_{8})$$
Using that $SL(2,3)$ and therefore $GL(2,3)$ have $4$ Sylow-$3$ subgroups one get a homomorphism into $Sym(4)$. It is straight forward to show that the kernel of this homomorphism is $Z(GL(2,3))$ and therefore $PGL(2,3)$ is isomorphic to a subgroup of $Sym(4)$. Since the two groups have the same number of elements we are done.
A: 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 $m\geq 2$.
