Confusion about meaning of 1, 1', and 2 in finding $S_3$ irreps, and how they correspond to the representation of the group elements I am working through the paper Non-Abelian Discrete Symmetries in Particle Physics, which can be found here. I am confused about the following highlighted passage found on pg. 8:

I understand this follows from the fact that there are two one dimensional irreps and one two dimensional irreps. My main question is asking if my following interpretation of this is correct: 
The above statement is saying that the representation of each group $g\in G$ element can be written as the direct sum 
\begin{align}
D(g)=\mathbf{1}_{g}\oplus\mathbf{1'}_g\oplus \mathbf{2}_g.
\end{align}
I don't think this is correct, but I'm a bit stuck at the moment. Could somebody please clarify how I go about finding $\mathbf{1}, \mathbf{1'}$, and $\mathbf{2}$, and how exactly that corresponds to the representation of the group elements?
 A: Some context might be helpful:
In the paper you refer to, $m_n$ denotes the number of $n$-dimensional irreps of $S_3$. The passage you are reading to is trying to count how many irreps there are of each dimension.
The paper quotes two well-known facts:


*

*The order of the group equals $m_1 + 4m_2 + 9m_3 + 16m_4 + \dots $ 

*The number of conjugacy classes equals $m_1 + m_2 + m_3 + m_4 + \dots$
Since the order of $S_3$ is $6$, and the number of conjugacy classes is $3$, the only possibility is that
$$ m_1 = 2, \ \  \ m_2 = 1, \ \  \ m_3 = m_4 = \dots = 0.$$
The irreps of $S_3$ are:


*

*The one-dimensional "identity" irrep $\mathbf{1}$, where every element is represented by $1$.

*The one-dimensioanl "alternating" irrep $\mathbf{1}'$, where every even element is represented by $1$ and every odd element is represented by $-1$.

*The two-dimensional irrep $\mathbf{2}$, obtained by viewing elements of $S_3$ as $2\times 2$ matrices describing symmetries of a triangle.


Your statement that "the representation of each group element $g \in G$ can be written as the direct sum $\mathbf{1}_g \oplus \mathbf{1}'_g \oplus \mathbf{2}_g$" doesn't make a great deal of sense to me, I'm afraid! There is no such thing as "the representation". There are many possible representations of $S_3$: any direct sum of $\mathbf{1}$, $\mathbf{1}'$ and $\mathbf{2}$ will do.
