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Consider the transpose inverse automorphism on $GL_n(\mathbb F)$ where $n\geq2$ and $|\mathbb F|>2$. (i.e. $\mathbb F$ is a field, possibly infinite, with three or more elements). I want to show this automorphism is not inner. I was told to consider $\det(BAB^{-1}) = \det(A)$ and $\det(\,^TA^{-1})=\det(A)^{-1}$ and derive a contradiction.

However, in some fields (where the result holds) I fail to see the contradiction. What about $\mathbb F_3$? An element of $GL_2(\mathbb F_3)$ either has determinant $1$ or $2$, both of which are their own multiplicative inverses in $\mathbb F_3$. I'm not sure if my knowledge of fields and determinants is flawed or the hint is.

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I think you are right, the hint does not work when $|F|=3$. An alternative proof that works for $n \ge 3$ is that inner automorphisms map stabilizers of subspaces of dimension $m$ to other such stabilizers, whereas inverse-transpose (the duality map) maps them to stabilizers of subspaces of dimension $n-m$. This argument works even for $|F|=2$.

When $|F|=n=2$, inverse-transpose really is inner. That leaves $n=2$ and $|F|=3$. In that case, inverse-transpose maps $\left(\begin{array}{cc}1&1\\1&0 \end{array}\right)$ to $\left(\begin{array}{cc}0&1\\1&2 \end{array}\right)$, which has a different trace, so it can't be inner.

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  • $\begingroup$ Now I'm more troubled than to begin with P: why does the first hint not apply to $n=2$ as it seems to me like it should? When $F$ is finite, cant you even replace the word dimension with the word cardinality? $\endgroup$ Dec 29, 2018 at 9:46
  • $\begingroup$ Sorry but I cannot make sense of either of your questions! The OP correctly observed that the hint given did not work in the case $|{\mathbb F}|=3$, and I was supplying alternative arguments for that case. The case $|{\mathbb F}|=2$ was excluded from the original question, but in fact the result is true in that case when $n \ge3$ but false when $n=2$. $\endgroup$
    – Derek Holt
    Dec 29, 2018 at 12:49

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