I was reading this proof and the author claimed that:

all inner automorphisms [of $D_4$] have order 2

How to prove it? Is it right for inner automorphisms of other group?

  • $\begingroup$ It is not true in general. Consider conjugation by $(1,2,3,\ldots,n)$ in $S_n$ It is true in that proof, which can be seen by checking each one. $\endgroup$ – Callus - Reinstate Monica Oct 20 '17 at 11:54
  • $\begingroup$ Thank you! But I'm wondering that do I have to check each one? I mean, if there is some more abstract way to prove it? $\endgroup$ – Ping Wan Oct 20 '17 at 13:04
  • $\begingroup$ There is no way to prove it, because you cannot prove a wrong statement. We have $Inn(G)=G/Z(G)$. $\endgroup$ – Dietrich Burde Oct 20 '17 at 13:08

We have that $\operatorname{Inn}(D_4) \cong D_4/Z(D_4)$, so it suffices to show every element of $D_4/Z(D_4)$ has order two. But every element of $D_4$ has order at most two, except $r$ and $r^3$. So we have reduced the problem to checking the order of $rZ(D_4)$ and $r^3Z(D_4)$ in $D_4/Z(D_4)$. And since $Z(D_4) = \{ e , r^2 \}$ we are done.

Alternatively you could verify each inner automorphism individually. To this end, let $\sigma_g \in \operatorname{Inn}(D_4)$, where $\sigma_g(x) = g^{-1}xg$ . Considering the square of this map yields, $$(\sigma_g)^2(x) = \sigma( \sigma (x) ) = g^{-2}xg^2.$$

Again we use the fact that $g^2 \in Z(D_4)$, for all $g \in D_4$. That is $(\sigma_g)^2(x) = x$, for all $x \in D_4$. And we have shown that $\sigma_g$ has order $2$.

Finally, it is not true in general that elements $\operatorname{Inn}(G)$ always have order 2. For example, $\operatorname{Inn}(D_3) \cong D_3$ because $D_3$ is centreless.


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