Groups of order 21 Let $G$ be a group of order 21:
1) Determine all possible values of $n$, where $n$ is the number of conjugacy classes of $G$.
2) Determine all the possible decomposition of $\mathbb{C}[G]$ as a product of matrix rings.
Could you help me solve these 2 problems?
EDIT: I think that to solve 1 maybe we can use the fact that the number of conjugancy class of $G$ is equal to the number of distinct simple $\mathbb{C}[G]$-modules.
 A: While Chris Eagle's answer is probably the preferred one, you can do this using rep theory, which it seems like might be how you want to do it.
1) We know that $\displaystyle 21=\sum_{\chi\in\text{irr}(G)}\chi(1)^2$. Now, we know that each $\chi(1)\mid 21$ and so $\chi(1)=1,3,7,21$, but since $\chi(1)^2<21$ we may conclude that $\chi(1)=1,3$. So, we have the equation $m_1+9m_2=21$. But, we know that $m_1=|G^\text{ab}|$ so that $m_1\mid 21$ so that $m_1=1,3,7,21$. I think you can deduce from this that the only solutions are $(m_1,m_2)=(21,0)$ and $(m_1,m_2)=(3,2)$. This tells you that there are either $21$ or $5$ conjugacy classes. Moreover, you know that both exist since not all groups of order $21$ are abelian (why?)
2) From the previous analysis I think you can easily conclude that either, as $\mathbb{C}$-algebras,
$$\mathbb{C}G\cong\mathbb{C}^{21}$$
or
$$\mathbb{C}G\cong\mathbb{C}^{3}\times\text{Mat}_3(\mathbb{C})^2$$
Depending on whether $G$ is abelian or not.
EDIT: Let $\widehat{G}_\mathfrak{L}$ denote the set of all linear characters with group operation being tensor product. Show then that $\widehat{G}_\mathfrak{L}=\text{Hom}_\mathbf{Grp}(G,\mathbb{C}^\times)$. But, since $\mathbb{C}^\times$ is abelian we have that $\text{Hom}_\mathbf{Grp}(G,\mathbb{C}^\times)$ is equipotent to $\text{Hom}_\mathbb{Z}(G^\text{ab},\mathbb{C}^\times)$. That said, by earlier comment this is equal to $\widehat{G^{\text{ab}}}_\mathfrak{L}$, but since $G^\text{ab}$ is abelian all its characters are linear, and it has $|G^{\text{ab}}|$ characters. The conclusion follows.
