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I proved that the mapping $\phi: G \to \operatorname{Aut}(G)$ that sends $g \longmapsto c_g (x) = gxg^{-1}$ is a homomorphism, but I need to find a group $G$ for which this map establishes an isomorphism. I know that $S_3$ is isomorphic to $\operatorname{Aut}(G)$, but I don't know off-hand whether this bijection is the correct one. I referenced a similar answer here Proving that ${\rm Aut}(S_3)$ is isomorphic to $S_3$, but it doesn't give an explicit formula for the bijection, and though I understand that $S_3$ is generated by its transpositions and so any automorphism is determined completely by how it acts on $(12)$, $(13)$, and $(23)$, I don't understand how a bijection follows.

Updated Attempt:

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. I claim that $\phi$ is also surjective. We have $S_3 = \langle a = (12), b = (13), c = (23) \rangle$, each of which has order $2$, so given $g \in S_3$, we have $g = a^i b^j c^k$ for $i,j,k \in \{0,1,2\}$. Since an automorphism $f \in Aut(S_3)$ 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, there are exactly $6$ automorphisms, so $|S_3| = |\mathrm{Aut}(S_3)|$. 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.

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  • $\begingroup$ General thought: (If I'm not mistaken) If for every element in $G$ (excluding $e$) there is an element that does not commute with it, this question is equivalent to asking if there are no outer automorphisms. Maybe that's one place to look? $\endgroup$ – Cameron Williams Mar 10 at 2:53
  • $\begingroup$ Hint: What is the kernel of the map $G\to \operatorname{Aut}(G)$? Also show that the fact automorphisms are determined by their actions on $3$ elements means that there is an embedding $\operatorname{Aut}(S_3)\to S_3$. $\endgroup$ – jgon Mar 10 at 3:05
  • $\begingroup$ The kernel is the set of $g \in G$ such that $\phi$ sends $g$ to the identity map $c_e (x) = xex^{-1} = e$. I'm not sure how to prove in closed-form that the only such $g$ is the identity element, however. It's certainly the case that two-cycles are stable under conjugation by two cycles, but not necessary (I believe) by three cycles. Could you explain a bit more about how there is an implied embedding to $S_3$? $\endgroup$ – user861776 Mar 10 at 3:19
  • $\begingroup$ Yes, this condition is equivalent to the one that $Z(G)=1$ and $Out(G)=1$. There are many examples of such groups, $S_3$ is one of them. (Actually, every $S_n$ when $n\ne 2, 6$.) See for instance here. $\endgroup$ – Moishe Kohan Mar 10 at 3:34
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    $\begingroup$ Just to clarify earlier comments: the conditions $Z(G)=1$ and ${\rm Out}(G) = 1$ together imply that $G \cong {\rm Aut}(G)$, and that proof works for $G=S_3$, but the converse is false. The group $D_8$ (dihedral of order 8) satisfies ${\rm Aut}(G) \cong G$, but $Z(G) \ne 1$. $\endgroup$ – Derek Holt Mar 10 at 8:03
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Your map $\phi:G\to Aut(G)$ has image $im(\phi)=Inn(G)=\{c_g| \ g\in G\}$ and $Aut(G)/Inn(G)=Out(G)$ so as mentioned in the comments, $\phi$ is onto iff $Out(G)=1$. Also $ker(\phi)=Z(G)$ hence $\phi$ is $1-1$ iff $Z(G)=1$. To sum up, $\phi$ is an isomorphism iff $Out(G)=Z(G)=1$.

It could be $Aut(G)\cong G$ with the isomorphism between $G$ and $Aut(G)$ being othen than $\phi$.

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  • $\begingroup$ Thanks for the answer. I haven't studied Out (G) yet, so I didn't allude to it in my solution. Could you tell me if my attempt is incorrect? $\endgroup$ – user861776 Mar 10 at 17:40

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