Character table of $S_3 \times C_2$ 
I need get of character table of $S_3 \times C_2$. How to make this character table?

The representation is a $\psi (g,h) = \rho_1 (g) \rho_2 (h)$ with $\deg (\rho _2) = 1$ and $\rho _1 $ irreducible.
Someone can help me?
Thanks in advance.
 A: Rather than just copying out the character table verbatim, here is a rough walkthrough on how to construct it.
It's quite hard to know what kind of theorems, facts, or methods you already know, so if there's anything that seems unfamiliar to you just let me know and I'll try to rephrase it!

*

*The character table of $G=S_3\times C_2$ can be found very easily from the character tables of $H_1=S_3$ and $H_2=C_2$ individually: irreducible characters of the product are exactly products of irreducible characters of the factors.
That is, each conjugacy class of $G$ has a representative $g=(h_1,h_2)$ for some $h_i\in H_i$. Say that $H_i$ has irreducible characters $\chi_j^{(i)}$. Then the values of the irreducible characters of $G$ on the conjugacy class of $g$ are exactly $\chi_j^{(1)}(h_1)\chi_k^{(2)}(h_2)$ for all possible values of $j,k$.


*To find the character table of $C_2$, note that, since the degrees of all irreducible characters must sum to the size of the group, we must have exactly two irreducible characters, both linear (degree 1). It is a fact (that you might already know) that linear characters are group homomorphisms. We already have the trivial character (which takes value $1$ on all elements of $C_2$), and then our other character must take the value $1$ on the identity element of $C_2$, and then $-1$ on the other element, since we need a number which forms an order-2 multiplicative group inside $\mathbb{C}$, with identity $1$, and $-1$ is the only other square root of unity.


*To find the character table of $S_3$ we have the trivial character, and then can lift any irreducible character of $S_3/S_3'$ to obtain a linear character of $S_3$. Now $S_3'\cong A_3\cong C_3$, and $S_3/S_3'\cong C_2$. But we already know the irreducible characters of $C_2$ from the above!
Lifting the trivial character just gives us the trivial character again, and lifting the only other character gives us a linear character  of $S_3$ corresponding to the parity (odd/even permutations) of an element.
Now we have two linear characters, and a natural place to look for irreducible characters of $S_n$ is the permutation character, defined by $\pi(g)=|\mathrm{fix}(g)|-1$, where $\mathrm{fix}(g)=\{1\leqslant i\leqslant n \mid gi=i \}$.
It turns out here that this character is irreducible (we can check this by showing that $\langle\pi,\pi\rangle=1$, for example) and is of degree $2$.
Then, since $1^2+1^2+2^2=6=|S_3|$, we are done.
Edit: for the sake of completeness, here are the character tables of $S_3$, $C_2$, and $S_3\times C_2$, taken from the (excellent) book Representations and Characters of Groups by James and Liebeck.

