3

What are the possible dimensions of an irreducible representation of a group of order 21?

I'm not sure what theorems might be useful for solving this problem. I know that every element of the group has order either 1, 3, or 7, and that there is at least one element of order 3 and one of order 7 (because not every element can have order p where $p\in \{21,3,7\}$, there is an identity element, and there is an element of order either 3 or 7). The dimension of the representation is by definition $\chi(1)$ where $\chi$ is the character of the representation. I know that the group is completely reducible as it is finite, though I'm not sure if this helps.

Deif
  • 1,002
user1127
  • 471
  • 8
  • 1
    Threre could be elements of order $21$, not just orders $1$, $3$, or $7$... – Arturo Magidin Mar 12 '24 at 16:36
  • 3
    Do you know the relations between the dimensions of irreducible representations and the order of a group, in general? – Captain Lama Mar 12 '24 at 16:50
  • 1
    What constraints do you know about? For example, if you know that the sum of the squares of the dimensions of the irreps. is $21$, then you know that the dimension of any irrep is $\leq 4$. It's a standard exercise to show that there are just $2$ groups of order $p q$, where $p, q$ are primes and $p \mid (q - 1)$, as is the case here ($p = 3, q = 7$), namely $C_{pq}$ and $G := C_q \rightthreetimes C_p$. It's not too hard to work out the character table of $G$ explicitly: https://math.stackexchange.com/questions/970660/character-table-of-the-non-abelian-group-of-order-21 – Travis Willse Mar 12 '24 at 16:58

1 Answers1

2

I shall talk in terms of character degrees. We denote by $\text{Irr}(G)$ the set of irreducible characters of $G$. I will prove that, indeed, the irreducible characters of $G$ are all linear or $G$ has $3$ linear characters and $2$ irreducible characters of degree $3$.

If $G$ is abelian then necessarily $\chi(1)=1$ for each $\chi\in\text{Irr}(G)$.

If $G$ is not abelian, since $\chi(1)$ divides $|G|$ for every $\chi\in\text{Irr}(G)$ and $\sum_{\chi\in\text{Irr}(G)}\chi(1)^2=|G|=21$ then necessarily $\chi(1)\in\{1,3\}$. In this case, there exists at least one $\chi\in\text{Irr}(G)$ with $\chi(1)=3$, as $G$ is nonabelian. The complex conjugate of an irreducible character is an irreducible character. Hence, $\overline\chi\in\text{Irr}(G)$. To prove the statement it suffices to check that $\chi\not=\overline\chi$.

Suppose that $\ker\chi\not=1$. Then it holds that $G/\ker\chi$ is a group of prime order. Now, since $\chi\in\text{Irr}(G/\ker\chi)$ and $G/\ker\chi$ is abelian it follows that $\chi(1)=1$, which is a contradiction. Thus, $\ker\chi=1$.

If $\chi=\overline\chi$ then for every $g\in G$ it would hold $$\chi(g)=\overline\chi(g)=\chi(g^{-1}).$$ Since $\ker\chi=1$, this yields that $g=g^{-1}$ and hence $g^2=1$ for all $g\in G$. But this implies that $G$ is abelian, which is a contradiction. Hence $\chi\not=\overline\chi$ and we are done.

Deif
  • 1,002
  • 1
    Great answer, but I think you can shorten it a bit, since the number of linear characters is equal to $[G:G']$, and it must be that $$|G|-[G:G']\equiv0\pmod{9}$$ – Steve D Mar 13 '24 at 00:45
  • @SteveD Thanks! You're right. $\sum_{\chi\in\text{Irr}(G)}\chi(1)^2$ and $G$ nonabelian forces that $|G:G'|=3$. You should submit that answer. – Deif Mar 15 '24 at 11:38