# Answering questions from a Character Table

I have been studying a course on Representation Theory, specifically Character Tables. I am able to construct a character table, but I cannot seem to understand working backwards with receiving the table. The question on a past paper exam is:

Consider the character table below of the finite group $G$.

where $i\in\mathbb{C}$ with $i^2=−1.$

(a) Determine $|G|$ and $|C_G(g_i)|$ for $i=1,...,6$.

(b) Show that $G/[G, G]$ is cyclic of order 4.

(c) Deduce that $G$ has a normal Sylow 3-subgroup $P$.

(d) Show that $C_P(g_5)=1$.

(e) Deduce that $P$ isomorphic to $C_3\times C_3$.

Any help is much appreciated!

• Which parts can you do? – Oscar Cunningham Apr 20 '17 at 15:06
• I can do part (a) and some of (b) (though I don't think my answer is right for (b)). I don't know where to begin for (c) onwards. – TejIhI Apr 20 '17 at 15:07
• @TejIhI It would be nice if you show your work on (a), (b) and (c), so that others can provide better help. – lisyarus Apr 20 '17 at 15:10
• I suspect that there is a typo and $\chi_4(g_5) = -i$. – Alex Provost Apr 20 '17 at 16:02

I'll post a more-or-less complete answer instead of giving hints because I think there are some mistakes in the question and so only giving hints would probably lead to more confusion.

(a) We know that the dot product of each column with itself is $|C_G(g_i)|$ . So for $i=1,\dots,6$ we have $36,4,9,9,4,4$. So clearly $g_1=e$ and $|G|=36$. The sizes of the conjugacy classes are $|G|/|C_G(g_i)|$ which gives $1,9,4,4,9,9$. Note that these do add to $36$.

(b) The group $G/[G,G]$ is called the abelianization of $G$. It is an abelian group with the special property that any homomorphism from $G$ to an abelian group factors through the quotient map $G\to G/[G,G]$. Given any $1$-dim rep $\chi:G/[G,G]\to \mathrm{GL}(V)$ we get a $1$-dim rep of $G$ by composing $G\to G/[G,G]\to \mathrm{GL}(V)$. It's easy to prove that distinct reps of $G/[G,G]$ give distinct reps of $G$ (using the fact that the quotient map is surjective). Finally if $V$ is $1$-dim then $\mathrm{GL}(V)$ is abelian, so any $1$-dim rep of $G$ factors through $G/[G,G]$. So the $1$-dim reps of $G$ arise uniquely from the reps of $G/[G,G]$ which are all $1$-dim because $G/[G,G]$ is abelian. Since abelian groups have the same number of reps as their order, we conclude that $|G/[G,G]|=4$ and since the $\chi_3$ and $\chi_4$ reps are complex they must arise from $C_4$ and not $C_2\times C_2$.

By the way, I strongly suspect the table is wrong and $\chi_4(g_5)=-i$. Even then the last two columns don't look orthogonal to me.

(c) Since $|G/[G,G]|=4$ we have $|[G,G]|=9$. Let $P=[G,G]$. The definition of a Sylow subgroup is just one whose order is the largest power of some prime dividing the order of the group, so this is a Sylow $3$-group. The commutator subgroup is always normal, which is how we could quotient by it in part (b).

(d) I'd been assuming above that $C_G(-)$ meant "centralizer" but the centralizer of $g_5$ in $P$ can't be $1$ because in part (e) we're going to see that $P$ is abelian so in $P$ everything will be in the centrailzer of everything else. In fact $g_5$ isn't even in $P$. The map $G\to G/[G,G]$ sends everything in $P$ to $e$, so by what we said in part (b) everything in $P$ must have character $1$ for each of the $1$-dim reps. This means that each element of $P$ must be in the conjugacy class of $g_1$, $g_3$ or $g_4$. From (a) these conjugacy classes only have $9$ elements between them, so $P$ is formed from precisely all the elements of these conjugacy classes.

Added: AlexProvost suggests $C_P(g_5)$ means $C_G(g_5)\cap P$, in which case it's obvious that the intersection of these two groups must be $\{e\}$ since the order of any element in their intersection must be a factor of both $4$ and $9$.

(e) Since $P$ has order $9$ it must be $C_9$ or $C_3\times C_3$. By restriction of the reps $\chi_1$, $\chi_5$ and $\chi_6$ we know some characters.

$$\begin{matrix} &g_1&g_2&g_3&g_4&g_5&g_6&g_7&g_8&g_9\\ \hline \chi_1'&1&1&1&1&1&1&1&1&1\\ \chi_5'&4&1&1&1&1&-2&-2&-2&-2\\ \chi_6'&4&-2&-2&-2&-2&1&1&1&1\\ \end{matrix}$$ These have inner products $<\chi_1,\chi_5>=<\chi_1,\chi_6>=<\chi_5,\chi_6>=0$, $<\chi_5,\chi_5>=4$ and $<\chi_6,\chi_6>=4$. So each of $\chi_5$ and $\chi_6$ must be the sum of $4$ distinct reps, with no overlap with each other or the trivial rep. The group $C_9$ has character table

$$\begin{matrix} &g_1&g_2'&g_3'&g_4'&g_5'&g_6'&g_7'&g_8'&g_9'\\ \hline \chi_1&1&1&1&1&1&1&1&1&1\\ \chi_2&1&\omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6&\omega^7&\omega^8\\ \dots&&&&&\dots&&&&\end{matrix}$$

where $\omega^9=1$ and the rest of the rows are also made of powers of $\omega$. I suspect it's not too hard to show that $-2$ can't be written as the sum of four distinct powers of $\omega$, and hence $P$ is not $C_9$. If you can't make that strategy work, I suspect the easiest proof technique will still be "prove $P\neq C_9$" rather than "prove $P=C_3\times C_3$".

Added: Okay, we can solve (e) with (d). Since $g_5$ commutes only with the identity on $P$, the automorphism $\alpha:p\mapsto g_5pg_5^{-1}$ of $P$ fixes only the identity (it fixes $P$ since $P$ is normal). Using exactly the same logic the same can be said of $g_2$. Since $\chi_3$ is $1$-dim we have $\chi_3(g_5^2)=\chi_3(g_5)^2=-1$ so $g_5^2$ must be in the same conjugacy class as $g_2$. So not only does $\alpha$ have no non-trivial fixed points, neither does $\alpha^2$. Since $\chi_3(g_5)=i$, the image of $g_5$ in $G/[G,G]$ has order $4$. Furthermore the order of $g_5$ must divide into $4$ since it is a member of its own centralizer, which has order $4$. So $g_5$ has order exactly $4$, and hence so does $\alpha$. But the automorphism group of $C_9$ can be easily checked to be $C_6$, which has no automorphisms of order $4$. So $P$ must be $C_3\times C_3$ instead.

• The last two columns are orthogonal then: $1\cdot 1 + (-1) \cdot (-1) + i \cdot i + (-i) \cdot (-i) = 0$. Also, for d), I think we can just argue that $C_P(g_5) = P \cap C_G(g_5)$; and since the former has order $9$ and the latter $4$, their intersection must be trivial. – Alex Provost Apr 20 '17 at 16:35
• Thank you so much for the detailed solution. It will help me a lot with revision! – TejIhI Apr 21 '17 at 9:54