The order of $G$ is its number of elements.
It is cyclic if and only if there exists $g\in G$ such that $G=\langle g\rangle$, the cyclic subgroup generated by $g$.
It is abelian if and only if $xy=yx$ for all $x,y\in G$.
It is solvable if and only if there is a series of subgroups \[ \langle e\rangle = G_0 \leq G_1 \leq \cdots \leq G_n=G\] such that each $G_i$ is normal in $G_{i+1}$ and all quotients $G_{i+1}/G_i$ are abelian.
The character table shows values of all characters of irreducible complex representations of $G$. The rows are indexed by characters, and the columns by conjugacy classes of $G$.
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- Last edited by Kiran S. Kedlaya on 2019-05-03 00:04:30
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