From the character table of $A_4$, I understand the following Kronecker products hold: \begin{align} {\bf 3} \times {\bf 3} = {\bf 1} + {\bf 1}_1 + {\bf 1}_2 + {\bf 3} + {\bf 3}. \end{align}
However, a physics textbook which I use also suddenly says it can be decomposed into its symmetric and antisymmetric parts as: \begin{align} ({\bf 3} \times {\bf 3})_{\rm sym} &= {\bf 1} + {\bf 1}_1 + {\bf 1}_2 + {\bf 3}, \\ ({\bf 3} \times {\bf 3})_{\rm antisym} &= {\bf 3}, \end{align} and actually I cannot understand what it does mean.
First, I guess the ${\bf 3}$ itself cannot be decomposed into those two parts because if we construct symmetric and antisymmetric matrices from one irreducible representation $M(g)$ (acts on a vector space $V$ and $g$ is a group element) by hand like \begin{align} M_{[\rm sym]} = \frac{1}{2}(M + M^t), ~~ M_{[\rm antisym]} = \frac{1}{2}(M - M^t), \end{align} these matrices do not satisfy the definiton of representation.
Hence I guess there is a way to construct two parts from two ${\bf 3}$s, but I do not understand it.
My question: How should I understand the above decomposition?
Also, I'm wondering why a ${\bf 6}$ representation does not appear here. Although it is not a irreducible representation of $A_4$, but if we divide the product space (which is the representation space of ${\bf 3} \times {\bf 3}$) into symmetric and anti-symmetric spaces, the dimensions become $6$ and $3$, respectively. In the representation theory of ${\rm SU}(3)$, there is a famous (but I do not fully understand) decomposition ${\bf 3} \times {\bf 3} = {\bf 6} + {\bf 3}^*$. Is there any relation with this and the above decomposition?
I'm sorry for the rambling question.