Mixed symmetry of rank $3$ tensor I have rank 3 tensor $T_{ijk}$ with following properties:
$T_{ijk}=T_{jik}$
$T_{ijk}=-T_{kji}$
Is it true that there is the only one tensor of rank 3 with those properties and it is $T_{ijk}=0$.
I'm starting from the following
$T_{ijk}=-T_{kij}=-T_{ikj}=T_{jki}=T_{kji}=-T_{ijk}$
$\Rightarrow 2T_{ijk}=0$
$\Rightarrow T_{ijk}=0$
Where am I making a mistake? Because the result is strange enough. Does this mean that from the point of view of Young diagrams, the hook is not interesting?
 A: The question and accepted answer here miss the main point, which is that for tensors of mixed symmetry, the indices themselves do not obey that symmetry, which is somewhat counterintuitive. This is explained in the comments by Michael Seifert here.
Tensors of mixed symmetry are generated by applying Young symmetrizers. In your case, you want to generate a tensor with symmetry
 1  2  3 
which means that to an arbitrary tensor $T_{ijk}$, you apply the symmetrizers
$$[e-(13)][e+(12)]T$$
where $e$ is the identity permutation, so that you get the desired tensor
$$T_{ijk}+T_{jik}-T_{kji}-T_{jki}$$
which is non-trivial.
For your method, you correctly arrived at the conclusion that the only tensor with $T_{ijk}=T_{jik}$ and $T_{ijk}=-T_{kji}$ is the zero tensor. It is not the right way to generate tensors of mixed symmetry, for the reasons mentioned at the beginning of this answer.
A: Your work seems fine. This little fact is sometimes known as the $S_3$-lemma, and it can be stated in a very general way: if $S$ is a non-empty set, $G$ is a (say, abelian) group without elements or order $2$, and $T: S\times S \times S \to G$ is symmetric in the first pair, and skew-symmetric on the second pair, then $T =0$.
$$T(x,y,z) = T(y,x,z)=-T(y,z,x)=-T(z,y,x) = T(z,x,y) = T(x,z,y) = -T(x,y,z).$$
The fact that tensors represent multilinear maps is irrelevant here. If $G$ is non-abelian, same proof works replacing minus signs with inverses (this is what would be understood as skew-symmetry).
