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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 $G$.

Knowl status:
  • Review status: beta
  • Last edited by John Jones on 2012-07-01 07:26:41.434000