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Show that ${\rm Inn}(S_n)$ isomorphic to $S_n$ for $n\ge3$.

To do this, if I define some isomorphic function say $\phi$, where $\phi: S_n \to{\rm Inn}(S_n)$, then show that $\phi$ is bijective (by showing injective and surjective), and a homomorphism, is this enough to prove the statement above?

I show it won't hold for $S_2$, but I'm not sure if my proof is strong enough to prove isomorphism for all $n\ge3$.

Shaun
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eddie
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2 Answers2

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The map $j$ that sends

$a\in S_n$

to

$f_a:S_n\to S_n$ defined by $f_a(x)=axa^{-1}$

is a group homomorphism from $S_n$ to $\text{Inn}(S_n)$.

The question is whether $j$ is bijective.

It is surjective, almost by definition.

For $n\geq3$, it is injective. Why?

$f_a=f_b$

$\implies\forall x\in S_n:axa^{-1}=bxb^{-1}$

$\implies\forall x\in S_n:x=a^{-1}\;b\;x\;b^{-1}\;a$

$\implies\forall x\in S_n:x=(b^{-1}a)^{-1}\;x\;b^{-1}a$

$\implies b^{-1}a$ commutes with every element of $S_n$

$\implies b^{-1}a=e$

$\implies a=b$.

Chris Sanders
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    The last but one implication holds iff $Z(S_n)$ is trivial, which is the case for $n\ge 3$ but it should be mentioned. –  Mar 24 '22 at 08:24
  • so just showing bijectivity and homomorphism is enough to prove it's true for all $n\ge3$? – eddie Mar 24 '22 at 15:08
  • why does $(b^{-1}a)$ commute with everything ? – eddie Mar 25 '22 at 18:12
  • @eddie, because this is what "$\forall x\in S_n:x=(b^{-1}a)^{-1};x;b^{-1}a$" means, which is the previous step in the implication sequence. –  Mar 25 '22 at 20:11
  • ...and so on backwards. –  Mar 25 '22 at 20:33
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Recall that for every group $G$, $\operatorname{Inn}(G)\cong G/Z(G)$, and that $Z(S_n)$ is trivial for $n\ge 3$ (see e.g. here, or many others alike). (But the answer to your question is, yes: the point is just to prove that $\phi$ is an isomorphism, and the proof of the injectivity requires what is shown in the link.)