I saw somewhere that $G$ abelian group of odd order then $x\to x^2$ is automorphism of $G$. But if $G=Z/3Z$ then $0\to 0, 1\to 1, 2\to 1$ @ManeeshNarayanan
Where am i wrong?
got it, i was using multiplication, even though $Z/3Z$ has only operation of addition
In the study of field theory and partial differential equations, a Toda field theory (named after Morikazu Toda) is derived from the following Lagrangian:
L
=
1
2
[
(
∂
ϕ
∂
t
...
consider the options $2$, $rank T=3$ and $rank T^2=1$. We know that $rank T^2\ge rank T + rank T-4 \implies rank T^2\ge 2 $. But $rank T^2=1$. So, (2) is not possible.
consider the options $2$, $rank T=3$ and $rank T^2=1$. We know that $rank T^2\ge rank T + rank T-4 \implies rank T^2\ge 2 $. But $rank T^2=1$. So, (2) is not possible.
