# 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 $$S_4$$.

• 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. Commented Sep 14, 2012 at 23:17
• @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? Commented Sep 14, 2012 at 23:23
• $i$ 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.
– user641
Commented Sep 14, 2012 at 23:34
• 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.
– user641
Commented Sep 14, 2012 at 23:38
• @SteveD Why don't you make it the answer? Commented Sep 15, 2012 at 2:22

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

• Could you explain the step (2)? I don't understand why we have a subgroup $K$ isomorphic to $S_3$. Commented Sep 15, 2012 at 14:13
• @Kato: nice question; edited the answer. Commented Sep 15, 2012 at 15:23
• I'm not sure how the action of conjugation being faithful implies that $Aut(Q_8)=V_4\rtimes K$. Commented Dec 2, 2019 at 2:14
• (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!] Commented Dec 4, 2019 at 4:28
• I think one needs to prove $\ker\Phi\cap K=1$. Commented Jul 12, 2020 at 8:36

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.

• "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? Commented Sep 15, 2012 at 13:55
• 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? Commented May 16, 2020 at 19:25

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.

• What is $N/C$ Lemma? Commented Sep 15, 2012 at 13:58
• If $H\leq G$, then $C_{G}(H)\vartriangleleft N_{G}(H)$ and moreover $N_{G}(H)/C_{G}(H)\hookrightarrow Aut(H)$. :) Commented Sep 15, 2012 at 14:50
• math.stackexchange.com/a/30379/583 has the proof for Inn being its own centralizer in this case. Commented Sep 15, 2012 at 18:45
• $\quad +1\quad \ddot\smile\quad$ Commented Mar 18, 2013 at 0:56
• 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. Commented Jul 11, 2020 at 5:30

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.

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.

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 $m\geq 2$.