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I am trying to create a character table for $\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $ and I am having some difficulties.

I know that the number of irreducible complex representations of a finite group G is equal to the number of it's distinct conjugacy classes. As $G=\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $ is finite, and in particular it has 6 conjugacy classes (which must be (0,0), (1,0), (0,1), (1,1), (2,0) and (2,1), right?) thus there must be 6 irreducible complex representations. Additionally, as G is abelian then all the irreducible complex representations are 1-dimensional. This add up nicely, as the number of such representations is then equal to the order of G.

Now, I am not not sure how to use the fact that they are 1-dimensional to create a character table, as I am not sure what the representations actually are. I have tried following an example, but I am a bit confused on how to do this. In the example the representations has been denoted $T_i$ but it is not quite clear to me one would obtain the $T_i$'s and then fill out the table in this case.

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  • $\begingroup$ You will get a representation by mapping a generator of the group - for example $(1,1)$- to any $6$-th root of unity. $\endgroup$ Commented Mar 11, 2021 at 12:36

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Notice that $G=\mathbb{Z}/6\mathbb{Z}$. Take a generator $g\in G$ (for example the class of $1$). Then on any irreducible representation $g$ acts as multiplication by a $6$th root of unity. The $6$ possible roots of unity give you the $6$ possible irreducible representations.

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