After consulting another post on this website, e.g., here Proving that ${\rm Aut}(S_3)$ is isomorphic to $S_3$, I've come up with a proof that $S_3 \simeq {\rm Aut}(S_3)$. Could someone look it over to tell me if there are any holes? I don't know if my proof of surjectivity is the best way to go about it. It almost feels like "cheating."

I claim that $\phi: S_3 \to \mathrm{Aut}(S_3)$ sending $g \longmapsto c_g (x) = gxg^{-1}$ is an isomorphism. In general, $\phi: G \to \mathrm{Aut}(G)$ sending $g \longmapsto c_g$ is a homomorphism, so it suffices to show that $\phi$ is a bijection. We have: \begin{align*} g \in \mathrm{ker}(\phi) & \iff \phi(g) = c_g (x) = x, \; \forall x \in G \\ & \iff gxg^{-1} = x, \; \forall x \in G \\ & \iff gx = xg, \; \forall x \in G \\ & \iff g \in Z(G) = \{e\}, \end{align*} so the kernel of $\phi$ is trivial, so $\phi$ is injective. Since $S_3$ has exactly $6$ elements which must be sent to distinct elements of the ${\rm Aut}(S_3)$, this implies that $|{\rm Aut}(S_3)| \leq 6$. I claim that $\phi$ is also surjective. We have $S_3 = \langle a = (12), b = (13), c = (23) \rangle$, each of which has order $2$. Since an automorphism $f \in{\rm Aut}(S_3)$ must preserve the order of elements, $f$ must send transpositions to transpositions. Furthermore, upon fixing where $f$ send these transpositions, the rest of the map is determined. Since there are $3!$ possibilities for where to send the permutations and these are the only possible automorphisms, we have $|{\rm Aut}(S_3)| \leq 6$. Hence, $|{\rm Aut}(S_3)| = 6$. But $\phi(S_3) \leq \text{Aut}(S_3)$, so that they have the same order immediately implies that $\phi(S_3) = \text{Aut}(S_3)$, so $\phi$ is surjective, hence bijective.

  • $\begingroup$ In fact, the statement holds for any positive integer other than $2$ or $6$ in place of $3$. Groups for which the conjugation map to the automorphism group is an isomorphism are called complete groups. $\endgroup$ Commented Mar 11, 2021 at 2:49
  • $\begingroup$ I think a simpler way to show $|{\rm Aut}(S_3)| \leq 6$ is to focus on where $g = (12)$ and $h = (123)$ can be sent. Let $\varphi$ be an automorphism of $S_3$. Since $\varphi$ preserves order, $\varphi(g)$ has at most 3 possibilities (the 2-cycles) and $\varphi(h)$ has at most 2 possibilities (the 3-cycles). An automorphism of $S_3$ is determined by its values on $g$ and $h$ (they generate $S_3$), so the possible automorphisms are at most the number of choices "2-cycle and 3-cycle". The number of those choices is $3 \cdot 2 = 6$, so $|{\rm Aut}(S_3)| \leq 6$. $\endgroup$
    – KCd
    Commented Mar 11, 2021 at 2:56
  • 1
    $\begingroup$ This looks correct to me; not necessarily the best way to do it, as Keith Conrad points out, but I was trying to point out how to push your original argument to conclusion. $\endgroup$ Commented Mar 11, 2021 at 16:19
  • 3
    $\begingroup$ There's a typo: the injectivity of $\phi$ implies that $|Aut(S_3)|\ge 6$, not $\le$. $\endgroup$
    – user870827
    Commented Mar 11, 2021 at 21:42
  • $\begingroup$ Thank you for catching this, and thank you KCd and Arturo for the comments. $\endgroup$
    – user861776
    Commented Mar 11, 2021 at 21:51

2 Answers 2


There are a few problems, but all are fixable.

  1. It is not clear that every element $g$ can be written as $a^ib^jc^k$ with $i,j,k\in\{0,1,2\}$. While every element is a product of factors, each of which is either $a$, $b$, or $c$, you would need to show that you can rewrite one in which they occur in the "wrong" order so that they show up in the right order. For example, $ba$ happens to be equal to $ac$ so you are fine, but that is not immediate and needs to be checked.

  2. The expressions for $g$ are clearly not unique, since you describe $8$ elements but $S_3$ only has six. So there are some repeats in there. You would need to show your maps are well-defined. You do not do so.

  3. It is not necessarily true that you are totally free to decide where $a$, $b$, and $c$ go. After all, $bab=c$, so if you know what happens to $a$ and $b$, then what happens to $c$ is forced. You need more than just deciding that you will map the transpositions arbitrarily: you are claiming all such maps are automorphisms, and you would need to check that. You do not do so. You do not even check that they are homomorphisms.

Now: you actually never use the fact that elements can be written in the form $a^ib^jc^k$, so why are you mentioning it? No need. In general, if $X$ generates $G$, then any group homomorphism is completely determined by what happens on $X$. This would take care of both 1 and 2: it doesn't matter whether you can write them in that specific way, and it doesn't matter that the expressions are not unique.

As for 3, you are asserting too much again. You already know that $\mathrm{Aut}(S_3)$ has at least six elements. Now, as you note, under any automorphism $\phi$, the three transpositions must be permuted (because $\phi$ sends elements of order $2$ to elements of order $2$, and because two different transpositions cannot be mapped to the same transpositions, so the restriction of $\phi$ to the set $\{(12),(23),(13)\}$ gives a permutation of that set). That means that there are at most $6$ homomorphisms, since, as noted above, what $\phi$ does to the transpositions completely determines $\phi$, and there are only six ways you can permute the elements of this set.

Thus, you know there are at least $6$ automorphisms, and at most $6$ automorphisms, so $|\mathrm{Aut}(S_3)|=6$, and so the map you have $S_6\to\mathrm{Aut}(S_6)$, which is an injective group homomorphism between finite groups of the same size must be an isomorphism.

  • $\begingroup$ This was extremely helpful. One question: do I know that Aut($S_3$) has "at least" six elements because I proved the map is injective, and the domain has $6$ elements? $\endgroup$
    – user861776
    Commented Mar 11, 2021 at 2:20
  • $\begingroup$ @user861776: Yes. $\endgroup$ Commented Mar 11, 2021 at 2:37
  • $\begingroup$ Thank you again. I edited my attempt in light of your suggestions. Would you mind taking another look at it? I want to be sure that the logic is as precise as possible. $\endgroup$
    – user861776
    Commented Mar 11, 2021 at 2:45

In general, given a group $G$, the group $\operatorname{Aut}(G)$ acts on the set of the conjugacy classes of $G$, $\mathscr{C}:=\{\operatorname{cl}(a), a\in G\}$, via $\varphi\cdot \operatorname{cl}(a):=\varphi(\operatorname{cl}(a))$. Essentially, the proof that this is indeed an action boils down to proving that $\varphi(\operatorname{cl}(a))=\operatorname{cl}(\varphi(a))$. Since inner automorphisms fix each conjugacy class, we have that $\operatorname{Inn}(G)\le\operatorname{ker}\phi$, where $\phi\colon\operatorname{Aut}(G)\to S_\mathscr{C}$ is the homomorphism equivalent to the said action.

Now, if $G=S_n$, then $\operatorname{Inn}(G)=\operatorname{ker}\phi$, because the stabilizer of the conjugacy class of the transpositions is precisely equal to $\operatorname{Inn}(S_n)$ (see e.g. here). Therefore (first homomorphism theorem), $\operatorname{Aut}(S_n)/\operatorname{Inn}(S_n)\cong\phi(\operatorname{Aut}(S_n))$. If distinct conjugacy classes are "tagged" by the distinct orders of their elements, then the action is trivial, because any automorphism (order-preserving) is then class-preserving, and finally $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$. This happens for $n=3$, being the three distinct conjugacy classes of $S_3$ made of elements of order $1$ (the identity), $2$ (the three $2$-cycles) and $3$ (the two $3$-cycles). Therefore $\operatorname{Aut}(S_3)=\operatorname{Inn}(S_3)\cong S_3$.

By relaxing to some extent the sufficient condition ensuring that all the automorphisms are class-preserving ("each conjugacy class, an order of its elements"), we can address $n=4,5$ cases. In fact, if $S_n$ has two conjugacy classes of elements of the same order, but whose (classes') sizes are different, than again all the automorphisms are class-preserving (being $|\varphi(\operatorname{cl}(\sigma))|=|\operatorname{cl}(\sigma)|$), and hence again the action is trivial (whence again $\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$). For $n=4$, the conjugacy classes of the transpositions and of the double transpositions (both whose elements have order $2$) have sizes $6$ and $3$, respectively, and hence no one automorphism can "swap" each other. Likewise, for $n=5$, the conjugacy classes of the transpositions and of the double transpositions have sizes $10$ and $15$, respectively, and the same conclusion as for $n=4$ does hold.

This simple approach can't support any further in the $n=6$ case, where we have conjugacy classes of elements of the same order, which (classes) have also the same size.


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