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Let $Q_8$ be the quaternion group.
How do we determine the automorphism group $Aut(Q_8)$ of $Q_8$ algebraically?
I searched for this problem on internet.
I found some geometric proofs that $Aut(Q_8)$ 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 $Aut(Q_8)$ is isomorphic to $S_4$.

$Q_8$ has three cyclic subgroups of order 4: $langle irangle$, $langle jrangle$, $langle krangle$, and $Aut(Q_8)$ acts on these three subgroups; inducing a homomorphism $Phicolon Aut(Q_8)rightarrow S_3$. We can see that, the homomorphism is surjective, since the two automorphisms $fcolon imapsto j, jmapsto i$, and $gcolon jmapsto k, kmapsto j$ give two transpositions in $S_3$.
The kernel contains those $varphiin Aut(G)$ such that $varphi(langle irangle)=langle irangle$ and $varphi(langle jrangle)=langle jrangle$ (automatically, $varphi(langle krangle)=langle krangle$).

(1) $varphi(langle irangle)=langle irangle$ 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:

$S: imapsto j, jmapsto i$, (hence $kmapsto -k$) and

$Tcolon jmapsto k, kmapsto j$ (hence $imapsto -i$);

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 $langle S,T rangle=Kleq Aut(Q_8)$, such that $Kcong S_3$ and $Phi(K)=S_3$. Also,

$ker(Phi) cap K=phi $.

Therefore, $Aut(Q_8)=ker(Phi)rtimes K cong V_4rtimes S_3$.

Consider an element of $ker(Phi)$:

$f:imapsto -i$, $jmapsto j$,

and two elements of $Kcong Im$:

$gcolon imapsto j, jmapsto i $ (like a transposition), and $hcolon imapsto j, jmapsto k$ (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 $V_4setminus{1}$ commutes with any element of $Ksetminus { 1}$ (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, $Aut(Q_8)=V_4rtimes Kcong S_4$

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 $phi(j)inlanglephi(i)rangle$, 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 $6cdot4=24$ automorphisms.

Let $alpha$ be any permutation on the elements $lbrace i,j,krbrace$ (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 $lbrace i,j,krbrace$ 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
$$ alpha(i)alpha(j)=pmalpha(k),$$
and so we see each permutation gives rise to an automorphism (after a suitable choice of sign).

Since the subgroups $langle irangle$,$langle jrangle$,$langle krangle$ 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 $Inn(Q_8)cong Q_8/Z(Q_8)cong C_2times C_2$. Thus $Aut(Q_8)$ has size at least $|S_3|cdot |C_2times C_2|=24$ automorphisms.

Since $Inn(Q_8)lhd Aut(Q_8)$, we get a semidirect product $(C_2times C_2)rtimes S_3$ 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.

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.

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 $mgeq 2$.

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.