I am trying to solve the character table and some related questions. The questions are below, and what I have done is below that. Any help on any pieces I am sure will enlightening. enter image description here

For parts c and d, I could really use some hints as I don't know where to begin.

e. The best I have figured out for the character table is

There are 5 conjugacy classes hence 5 irreducible representations. There 20 elements in the group and the only way to break this down to the sum of squares is $1^2 + 1^2 + 1^2 + 1^2 + 4^2$ so these are the orders of the conjugacy classes of the trivial element.

\begin{array}{rrrrrrrrrrr} & C_1 & C_2 & C_3 & C_4 & C_5\\ \chi_0 & 1 & 1 & 1 & 1 & 1\\ \chi_1 & 1& 1& i & -i & -1 & \\ \chi_2 & 1 & 1 & -i & -i & -1& \\ \chi_2 & 1 &1 &-1 & -1& 1 & \\ \chi_3 & 4 & -1 & 0& 0& 0& \\ C_G(x) & 1 & 5 & 4 & 4 & 4 & \end{array}

although I think $d,e$ should be a -i and i pair. With my calculations I am having a hard time just putting in numbers trying to satisfy $\left \langle \alpha, \beta\right \rangle := \frac{1}{|G|}\sum_{g \in G} \alpha(g) \overline{\beta(g)} $ and the column version of Schur's orthogonality relationship>

f. I see that the centralizer of $C_3$ has 4 elements but does this, and if so how, does this mean that elements $x \in C_3$ have order divisible by 4?

g. more properties of conjugacy classes and that somehow I can read this from the character table?

Really thanks for any hints, tips, or answers.


1 Answer 1


I am answering this question using Sylow's theorem as a crutch - I don't know whether that is 'allowed.' Still, I hope this helps.

Sylow says that the number 5-Sylow groups is a factor of $20/5$, and is congruent to $1 \pmod 5$. Therefore, there is a normal subgroup $S$ of order $5$, necessarily cyclic, in $G$, $S = C_1 \cup C_2$, and we have the exact sequence $$ 1 \rightarrow S \rightarrow G \rightarrow H \rightarrow 1,$$ where $H$ is a group of order $4$ and thus abelian. Therefore, there are $4$ 1-dimensional representations of $G$ which factor through $H$, i.e., the $4$ irreducible representations of $H$; since the table (completed by you) tells us that $i$ is one of the character values, we know that $H \simeq \mathbb Z /4Z$, and that gives four (multiplicative) characters: 1, $\chi$, $\chi^2$, and $\chi^3$, with $\chi(1) =\chi(C_2)=1$, $\chi(C_3) =i$, $\chi(C_4) =-i$, and $\chi(C_5) = -1$.

By Sylow again, the above sequence splits, i.e., $G \simeq S \rtimes H $. From the conjugacy class counts, in particular the cardinality of $C_2$, we know that $H$ maps onto ${\rm Aut}(S)$.

At this point, we know what $G$ is up to isomorphism: $G$ is isomorphic to the group of affine transformations $g_{{a,b}}\colon x \mapsto ax +b$, $a\in \def \F {{\mathbb F_5}} \F^*$ and $b\in \F$: $C_1 =1$, $g_{(1,1)}\in C_2$, $g_{(2,0)}\in C_3$, $g_{(3,0)}\in C_4$, and $g_{(4,0)}\in C_5$. (This deals with f and g of the question.) The commutator group $[G,G] = S$: on the one hand, $H$ is abelian, so $G^{ab}\rightarrow H$, and the commutator subgroup is a subgroup of $S$; on the other hand, $G$ is not abelian.

You already wrote down the 'missing' character (using orthogonality and arithmetic), but as a (less good) alternative: As $S\simeq \mathbb Z/5$, we know the 5 (multiplicative) characters of $S$, $\psi^k$, $k\in \mathbb Z/5$. The induced character $$ \Psi(g) = \sum_{k=1}^4 \psi' \left( g_{(k,0)}^{-1}\, g\, g_{(k,0)}\right),$$ where $\psi' (s) = \psi ( s)$ if $s\in S$ and $0$ otherwise, is the missing character: $\Psi( 1) = 4$, $\Psi ( C_2) = -1$ (because the sum of the non-trivial fifth roots of unity = $-1$ ), and zero otherwise.

  • $\begingroup$ I can not seem to figure out what you are talking about when you talk of the affine transformations in conjunction with parts f and g. Do you know of a simpler approach to tackle these parts of the question? Thanks much for your help, the first parts made some sense. $\endgroup$
    – Relative0
    Sep 19, 2015 at 20:58
  • $\begingroup$ Hi @Relative0, I'll follow up Monday at the latest - but as I remember, the 'affine transformations' bit is just a restatement that $G= S \rtimes H$. $\endgroup$
    – peter a g
    Sep 20, 2015 at 2:59
  • $\begingroup$ @Relative0 I started to write out a long spiel, and will continue if need be, but re-reading the question, I am not sure that I understand what is confusing you: do you agree that $G= S\rtimes H$, (my notation)? If so; that $H$ is a cyclic subgroup of order $4$, and its 3 non-trivial elements are representatives for $C_3$, $C_4$ and $C_5$? Hence the orders are $2$ or $4$? Lining them up with the char table info, and writing $H$ as the additive group $ \mathbb Z/4\mathbb Z$, we can choose $1 \in C_3$, $3 \in C_4$, and $2\in C_5$? $4\cdot 1 = 4 \cdot 3 = 0 $ and $2 \cdot 2 = 0$. (cont) $\endgroup$
    – peter a g
    Sep 22, 2015 at 2:29
  • $\begingroup$ (cont) (The affine transformation stuff is just notation.) $\endgroup$
    – peter a g
    Sep 22, 2015 at 2:31
  • $\begingroup$ Doing it your way - no you cannot conclude just from 4= size of the centralizers that elements of $C_3$ have order divisible by $4$ - after all, $C_5$ has the same property, but the orders are supposed to be $2$. But! From the char. table, there is a homomorphsim $\chi_1$ which takes $x\in C_3$ to $i$, which is an element of order $4$. In particular $4$ must divide the order of $x$... Also, from the table again, $x^{-1}\in C_4$, (because $x^{-1}$ has to belong some conjugacy class $C$, and $C_4$ is the only one with $ \chi_1(C)= -i$) and $x^2 \in C_5$ (same kind of reasoning). $\endgroup$
    – peter a g
    Sep 22, 2015 at 3:39

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