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 $S_4$.

  • $\begingroup$ Have a look at this: crazyproject.wordpress.com/2010/07/25/…. It basically boils down to the fact that no automorphism of $Q_8$ can have order 6. $\endgroup$ – M Turgeon Sep 14 '12 at 23:17
  • $\begingroup$ @MTurgeon Thanks. "We already know that $|Aut(Q_8)| = 24$ by counting the possible images of, say. $i$ and $j$". How do we know that? $\endgroup$ – Makoto Kato Sep 14 '12 at 23:23
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    $\begingroup$ $i$ can go to any of the $6$ elements of order $4$ in $Q_8$. After that, $j$ can go to any of four elements of order $4$; it can't end up in the same subgroup as $i$ (it won't be an automorphism), and since $i$ and $j$ generate $Q_8$, this is sufficient to determine the automorphism. $\endgroup$ – user641 Sep 14 '12 at 23:34
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    $\begingroup$ Now simply note that any permutation of $i,j,k$ gives an autmorphism. Each of these is outer, since the subgroups $\langle i\rangle$, $\langle j\rangle$, and $\langle k\rangle$ are normal. Thus the automorphism group contains the semidirect product of the inner automorphisms (of size 4), and the permutations on $i,j,k$ (of size 6). Now you are done. $\endgroup$ – user641 Sep 14 '12 at 23:38
  • $\begingroup$ @SteveD Why don't you make it the answer? $\endgroup$ – Makoto Kato Sep 15 '12 at 2:22

$Q_8$ has three cyclic subgroups of order 4: $\langle i\rangle$, $\langle j\rangle$, $\langle k\rangle$, and $Aut(Q_8)$ acts on these three subgroups; inducing a homomorphism $\Phi\colon Aut(Q_8)\rightarrow S_3$. 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 $\varphi\in Aut(G)$ 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 $\langle S,T \rangle=K\leq Aut(Q_8)$, such that $K\cong S_3$ and $\Phi(K)=S_3$. Also,

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

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

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, $Aut(Q_8)=V_4\rtimes K\cong S_4$

  • $\begingroup$ Could you explain the step (2)? I don't understand why we have a subgroup $K$ isomorphic to $S_3$. $\endgroup$ – Makoto Kato Sep 15 '12 at 14:13
  • $\begingroup$ @Kato: nice question; edited the answer. $\endgroup$ – Beginner Sep 15 '12 at 15:23

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.

  • $\begingroup$ "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)." I think the explanations are a bit too terse. Could you explain in more detail? $\endgroup$ – Makoto Kato Sep 15 '12 at 13:55


$Inn(Q_8)\cong V$ 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.

  • $\begingroup$ What is $N/C$ Lemma? $\endgroup$ – Makoto Kato Sep 15 '12 at 13:58
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    $\begingroup$ If $H\leq G$, then $C_{G}(H)\vartriangleleft N_{G}(H)$ and moreover $N_{G}(H)/C_{G}(H)\hookrightarrow Aut(H)$. :) $\endgroup$ – mrs Sep 15 '12 at 14:50
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    $\begingroup$ math.stackexchange.com/a/30379/583 has the proof for Inn being its own centralizer in this case. $\endgroup$ – Jack Schmidt Sep 15 '12 at 18:45
  • $\begingroup$ $\quad +1\quad \ddot\smile\quad$ $\endgroup$ – Namaste Mar 18 '13 at 0:56

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 $m\geq 2$.


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.


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