Show that the symmetric group of degree 3 $S_3$ has 6 automorphisms.

I can understand the argument that each transposition must permute. 3 transpositions gives $3! = 6$.

But the argument given in my textbook says the following.

$S_3$ can be represented as $\{1, \sigma, \sigma ^2, \tau, \tau\sigma, \tau\sigma ^2\}$ (i.e. isomorphic to the dihedral group of order 6), where $o(\sigma)=3$ and $o(\tau)=2$. Here I denote $o(x)$ to mean the order of the element $x$. So if $\theta$ is an automorphism then $o(\theta(\sigma))=o(\sigma)=3$, so $\theta(\sigma)=\sigma \text{ or } \sigma^2$ because $o(\sigma)=o(\sigma ^2)=2$.

I understand the argument up to here, but what follows I don't understand.

Similarly $\theta(\tau)=\tau, \tau\sigma \text{ or } \tau\sigma ^2$

They conclude that there are at most $2\cdot 3=6$ choices for $\theta$, which completes the solution based on a previous observation that $|\text{inn}S_3|=6$ and $\text{inn}S_3\subseteq\text{aut}S_3$.

My confusion is that I computed $o(\tau\sigma)$ and $o(\tau\sigma ^2)$ to be $6$, not $2$, whereas $o(\tau)=2$. Using this logic, there should be $1\cdot2\cdot1\cdot2=4$ automorphisms, where the identity element is sent to itself and $\tau$ is sent to itself.

It would really help me if somebody could list each of the 6 automorphisms so that I can clearly see why there are 6.

I apologize if this question has been answered before; I'm in a rush.

  • 2
    $\begingroup$ The orders of $\tau\sigma$ and $\tau\sigma^2$ are $2$. Could you share your computations? $\endgroup$ Nov 20, 2015 at 17:35
  • 1
    $\begingroup$ I computed the order of $\tau\sigma$ and $\tau\sigma^2$ incorrectly by assuming commutativity: e.g. $(\tau\sigma)^2=\tau^2\sigma^2=\sigma^2$. $\endgroup$
    – ahorn
    Nov 21, 2015 at 8:11

2 Answers 2


Note that $S_3$ is generated by $(1\ 2)$ and $(1\ 2\ 3)$: powers of either $(1\ 2)$ or $(1\ 2\ 3)$ account for $4$ elements of $S_3$ directly, and:

$(1\ 2\ 3)(1\ 2) = (1\ 3)$

$(1\ 2)(1\ 2\ 3) = (2\ 3)$

account for the remaining $2$.

So any homomorphism $\phi:S_3 \to S_3$ is completely determined by $\phi((1\ 2\ 3))$ and $\phi((1\ 2))$.

Now an automorphism, being an isomorphism, must preserve the order of elements. Thus we have only $2$ (at most) possible choices for what $\phi((1\ 2\ 3))$ can be, and only $3$ (at most) possible choices for what $\phi((1\ 2))$ might be. This gives at most $6\ (= 2\cdot 3)$ possible automorphisms. Since the inner automorphisms comprise $6$ distinct automorphisms, this must be all of them. Using $a = (1\ 2\ 3)$ and $b = (1\ 2)$, here are all $6$:

$1.\ \phi(a) = a, \phi(b) = b\\2.\ \phi(a) = a,\phi(b) = ab\\3.\ \phi(a) = a,\phi(b) = a^2b\\4.\ \phi(a) = a^2,\phi(b) = b\\ 5.\ \phi(a) = a^2,\phi(b) = ab\\6.\ \phi(a) = a^2,\phi(b) = a^2b$


By definition $S_3=\{f:\{1,2,3\}\to\{1,2,3\}\,|\, f \text{ is invertible}\}$. Let $\sigma:\{1,2,3\}\to\{1,2,3\}$ be the map defined by $\sigma(1)=2$, $\sigma(2)=3$, and $\sigma(3)=1$, and let $\tau:\{1,2,3\}\to\{1,2,3\}$ be map defined by $\tau(1)=2$, $\tau(2)=1$, $\tau(3)=3$. A straightfoward calculation shows that $|S_3|=6$, and hence if we find $6$ destinct maps we know we have found all of them. A tedious (but straightforward) calculation shows that $S_3=\{1,\sigma,\sigma^2,\tau,\tau\sigma,\tau\sigma^2\}$ where $1$ denotes the identity map, and furthermore that $\sigma^3=\tau^2=\sigma\tau\sigma\tau=1$.

Note that from this we can deduce that: $$\tau\sigma\tau\sigma=\tau\sigma\tau\sigma\tau\tau=\tau^2=1$$ $$\tau\sigma^2\tau\sigma^2=\tau\sigma\sigma\tau\sigma\tau\tau\sigma=\tau\sigma\tau\sigma=1$$ and hence $o(\tau\sigma)=o(\tau\sigma^2)=2$.

The six automorphisms turn out to be conjugation by each element in $S_3$. i.e $$x\mapsto x$$ $$x\mapsto \sigma x \sigma^2$$ $$x\mapsto \sigma^2 x \sigma$$ $$x\mapsto \tau x \tau$$ $$x\mapsto \tau \sigma x \tau \sigma$$ $$x\mapsto \tau\sigma^2 x \tau\sigma^2.$$

To understand why this is so note that for any group $G$ the group $G/Z(G)$ is isomorphic to the group of inner automorphisms of $G$, and hence since $S_3$ has trivial center each element determines a different automorphism.

  • $\begingroup$ Why is every automorphism an inner automorphism? $\endgroup$
    – user39082
    Apr 27, 2021 at 14:26
  • $\begingroup$ @user39082 I guess this could be a separate question. However, S3 is generated by $\sigma$ and $\tau$ above, hence an automorphism is determined by where these generates get sent. Since automorphisms preserve order and there are only 2 elements of order 3 (the order of $\sigma$) and 3 elements of order 2 (the order of $\tau$) it follows there are at most 6 automorphisms. $\endgroup$
    – Nex
    Oct 25, 2021 at 9:29

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