Are some irreducible representations redundant? I am very new to group theory so it is likely that this question could be very obvious.
I have just completed a question where we find the irreps of $D_4$, which the question defined as a group generated by the two matrices $$ \sigma = \begin{pmatrix} 1 &0 \\ 0 &-1 \end{pmatrix}, \ \ \epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
After finding the character table, I found that $D_4$ had 4 1D irreps (given below) and 1 2D irrep (just represented by the matrices generated by those above).
$$R_1 = \{1,1,1,1,1,1,1,1\}$$
$$R_2 = \{1,1,1,-1,-1,-1,-1,1\}$$
$$R_3 = \{1,-1,-1,1,1,-1,-1,1\}$$
$$R_4 = \{1,-1,-1,-1,-1,1,1,1\}$$.
My question is about the significance of these 5 irreps. It seems to me like the group can be completely represented by its 2D irrep (after all, in my question the group was defined just by these matrices) and that the 4 1D irreps provide no additional information. So firstly, am I correct in that these 1D irreps provide no additional information about the group and that it is uniquely defined by its 2D irrep?
Secondly, I am confused by the link between a group's irreps and its representation decomposition. Irreps are often described as the 'fundamental building blocks', but I'm not sure how exactly to interpret them here. Is the point that the group can be represented by a direct sum of these irreps, or a tensor product, or neither? If the former, it still seems like the 1D irreps provide no new information here.
(I have noticed that the -1's in the 1D irreps correspond to the size 2 conjugacy classes of 2D matrices which contain 2 matrices which are -1 $\times$ each other so maybe there is some information there?)
 A: There are two concepts here. The first is in what sense are the irreducible representations the building blocks of all representations. This, as mentioned in the comments, is because every representation over $\mathbb C$ is isomorphic to a direct sum of irreducible representations.
The second, which you are alluding to, is whether, given the $2$-dimensional representation, you need the others. What you seem to be driving at is that the $2$-dimensional representation is faithful, meaning the homomorphism to the matrix group is injective. Indeed, the $2$-dimensional representation is faithful, and any faithful representation of $D_4$ must have that $2$-dimensional representation as a summand. In that sense you do not need the other representations to define your group, and the other representations cannot be used, in that you always need the $2$-dimensional representation.
A finite group has a faithful, irreducible representation if and only if its centre is cyclic. So for the group $C_2\times C_2$ you cannot find a faithful irreducible representation, and you would need the sum of two of them to produce an isomorphic copy of your group. In general, if the centre needs $n$ elements to generate it, one needs $n$ irreducible representations to be faithful (and of course, you must choose your representations wisely to obtain a faithful representation).
