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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 $S_4$.

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  • $\begingroup$ 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
    Commented Sep 14, 2012 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? $\endgroup$ Commented Sep 14, 2012 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
    Commented Sep 14, 2012 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
    Commented Sep 14, 2012 at 23:38
  • $\begingroup$ @SteveD Why don't you make it the answer? $\endgroup$ Commented Sep 15, 2012 at 2:22

6 Answers 6

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

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  • $\begingroup$ Could you explain the step (2)? I don't understand why we have a subgroup $K$ isomorphic to $S_3$. $\endgroup$ Commented Sep 15, 2012 at 14:13
  • $\begingroup$ @Kato: nice question; edited the answer. $\endgroup$
    – Beginner
    Commented Sep 15, 2012 at 15:23
  • $\begingroup$ I'm not sure how the action of conjugation being faithful implies that $Aut(Q_8)=V_4\rtimes K$. $\endgroup$ Commented Dec 2, 2019 at 2:14
  • $\begingroup$ (1) If $N$ is Klein-$4$ group, (2) if $H$ is a group isomorphic to $S_3$, (3) And if $H$ acts on $N$ faithfully, then the semi-direct product $N\rtimes H$ is isomorphic to $S_4$. [The faithfulness is used to establish this isomorphism; after statement $\ker\Phi\cap K=\phi$, it is shown that ${\rm Aut}(Q_8)$ is semi-direct product of a Klein-4 group and a group isomorphic to $S_3$; so what is this semi-direct product? This was what I wanted to say!] $\endgroup$
    – Beginner
    Commented Dec 4, 2019 at 4:28
  • $\begingroup$ I think one needs to prove $\ker\Phi\cap K=1$. $\endgroup$
    – ashpool
    Commented Jul 12, 2020 at 8:36
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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.

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    $\begingroup$ "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? $\endgroup$ Commented Sep 15, 2012 at 13:55
  • $\begingroup$ If $\varphi(j) \in \langle \varphi(i) \rangle$, why is the map not surjective? Which element, for example, is not in the image of $\varphi$? Also, wouldn't the map fail to be injective in this case? $\endgroup$
    – Junglemath
    Commented May 16, 2020 at 19:25
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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.

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  • $\begingroup$ What is $N/C$ Lemma? $\endgroup$ Commented Sep 15, 2012 at 13:58
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    $\begingroup$ If $H\leq G$, then $C_{G}(H)\vartriangleleft N_{G}(H)$ and moreover $N_{G}(H)/C_{G}(H)\hookrightarrow Aut(H)$. :) $\endgroup$
    – Mikasa
    Commented Sep 15, 2012 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. $\endgroup$ Commented Sep 15, 2012 at 18:45
  • $\begingroup$ $\quad +1\quad \ddot\smile\quad$ $\endgroup$
    – amWhy
    Commented Mar 18, 2013 at 0:56
  • $\begingroup$ I don't understand how your hint implies that $\operatorname{Aut}(Q_8)\simeq S_4$. Could you elaborate? I'm sorry, I'm not a smart person. $\endgroup$
    – ashpool
    Commented Jul 11, 2020 at 5:30
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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.

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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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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 $m\geq 2$.

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