Lecture 16

Solution: We first note the relevant part of the character table of $A_{5}$:

Next we determine the conjugacy classes of $A_{6}$ and their representatives. Note that the only cycle type of distinct odd lengths is 5 + 1, that is, a 5-cycle. So this is the only conjugacy class of $S_{6}$ that ’splits’ in $A_{6}$.

The conjugacy classes and their sizes are given below:

We see that, for each conjugacy class ${\mathcal{C}}$ of $A_{6}$, ${\mathcal{C}}\cap A_{5}$ is either empty (if ${\mathcal{C}}$ is the class of $(123)(456)$ or $(1234)(56)$), or a single conjugacy class ${\mathcal{D}}$ of $A_{5}$. In the first case,

and in the second

We obtain

Solution: Write $\tilde{\chi}$ for the induced character. Since the only conjugacy classes of $S_{n}$ that intersect $H$ are the identity and the $p$-cycles, we see that

if $g$ is not a $p$-cycle or the identity. We have

Finally, let $g=(12\ldots p)$. Then the conjugacy class ${\mathcal{C}}$ of $g$ in $S_{n}$ contains all the nonidentity elements of $H$, each of which is its own conjugacy class in $H$ as $H$ is abelian. We therefore have

We have $|{\mathcal{C}}|=(p-1)!$. If $\chi(g)=\zeta$, then

by the hint. Therefore

This completes the solution to the first part. For the second part, if we compute $\langle\tilde{\chi},\tilde{\chi}\rangle$ we get

This is an integer (because $\tilde{\chi}$ is the character of a representation), so $(p-1)!+1$ is divisible by $p$, as required.

Solution:

1. (1)
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   Since $G$ is generated by $x$ and $y$, in order to show that $H$ is normal, it suffices to show that $xHx^{-1}=H$ and $yHy^{-1}=H$. Every element of $H$ is of the form $y^{j}$ for some $j\in{\mathbb{Z}}/7{\mathbb{Z}}$. We then have

   $$xy^{j}x^{-1}=y^{2j}\in H,\quad yy^{j}y^{-1}=y^{j}\in H,$$

   hence $H$ is normal.

2. (2)
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   We now determine the conjugacy classes of $G$. As usual, we choose an element, and successively conjugate it by the generators until we have a whole conjugacy class, and then repeat until every element of $G$ has been assigned a conjugacy class (cf. Problems 3,4,5):

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     $\{e\}$ is a conjugacy class

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     Taking $y$, conjugating with $y$ doesn’t do anything. Conjugating with $x$ gives $y^{2}$. Again, conjugating $y^{2}$ with $y$ does not give any new elements. Conjugating with $x$ gives $y^{4}$. Repeating gives $xy^{4}x^{-1}=y^{8}=y$, so $\{y,y^{2},y^{4}\}$ is a conjugacy class.

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     Starting with $y^{3}$ and conjugating with $x$ gives $y^{6}$. We then have $xy^{6}x^{-1}=y^{12}=y^{5}$, and $xy^{5}x^{-1}=y^{10}=y^{3}$, so $\{y^{3},y^{5},y^{6}\}$ is a conjugacy class.

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     Now taking $x$ and conjugating with $y$, we have (using $x^{-1}=x^{2}$)

     $$yxy^{-1}=x^{3}yx^{4}y^{-1}=x^{2}y^{2}x^{2}y^{-1}=xy^{4}y^{-1}=xy^{3}.$$

     Now conjugating $xy^{3}$ by $x$ gives

     $$x\cdot xy^{3}x^{-1}=xy^{6},$$

     and conjugating by $y$ gives

     $$yxy^{3}y^{-1}=x^{3}yx^{4}y^{2}=x^{2}y^{2}x^{2}y^{2}=xy^{4}y^{2}=xy^{6}.$$

     The conjugations of $xy^{6}$ give

     $$x^{2}y^{6}x^{-1}=xy^{12}=xy^{5},\quad yxy^{6}y^{-1}=x^{3}yx^{4}y^{5}=xy^{9}=xy^{2},$$

     and then conjugating $xy^{5}$:

     $$x^{2}y^{5}x^{-1}=xy^{10}=xy^{3},\quad yxy^{5}y^{-1}=x^{3}yx^{4}y^{4}=xy^{8}=xy.$$

     Conjugating $xy^{2}$ gives

     $$x^{2}y^{2}x^{-1}=xy^{4},\quad yxy^{2}y^{-1}=x^{3}yx^{4}y=xy^{5},$$

     conjugating $xy$ gives

     $$x^{2}yx^{-1}=xy^{2},\quad yxyy^{-1}=x^{3}yx^{4}=xy^{4},$$

     and finally conjugating $xy^{4}$ gives

     $$x^{2}y^{4}x^{-1}=xy^{8}=xy,\quad yxy^{4}y^{-1}=x^{3}yx^{4}y^{3}=xy^{7}=x,$$

     so $\{x,xy,xy^{2},xy^{3},xy^{4},xy^{5},xy^{6}\}$ is a conjugacy class.

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     Similarly to case iv), for any element of the form $x^{2}y^{j}$, we have

     $$y(x^{2}y^{j})y^{-1}=x^{3}yx^{2}y^{j-1}=x^{2}y^{j+1},$$

     hence all elements of the form $x^{2}y^{k}$ are in the same conjugacy class, i.e. the final conjugacy class is $\{x^{2},x^{2}y,x^{2}y^{2},x^{2}y^{3},x^{2}y^{4},x^{2}y^{5},x^{2}y^{6}\}$

   Since $|\widetilde{G}|=|G/H|=3$, $\widetilde{G}\cong C_{3}$, which has three irreducible representations: apart from the trivial representation, let $\widetilde{\psi}_{1},\widetilde{\psi}_{2}$ the one-dimensional be defined through

   $$\widetilde{\psi}_{i}(x^{j}N)=\omega^{ij}N,$$

   where $\omega=e^{2\pi i/3}$. Lifting these to $G$ gives three irreducible representations:

   size:	1	3	3	7	7
   	$e$	$y$	$y^{3}$	$x$	$x^{2}$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1	1
   $(\psi_{1},{\mathbb{C}})$	1	1	1	$\omega$	$\omega^{2}$
   $(\psi_{2},{\mathbb{C}})$	1	1	1	$\omega^{2}$	$\omega$

   We are now missing two irreducible representations of $G$, and by counting dimensions, we see that they both must have dimension three.

   Since $H\cong C_{7}$, it has seven irreducible representations, each of which we can induce to $G$. The character formula in our case reads

   $$\mathrm{Ind}_{H}^{G}\chi_{\pi}({\mathcal{C}})=\frac{3}{|{\mathcal{C}}|}\sum_{{\mathcal{D}}\subset{\mathcal{C}}}\chi_{\pi}({\mathcal{D}}),$$

   (where each ${\mathcal{D}}$ is a conjugacy class of $H$, and hence $|{\mathcal{D}}|=1$) so the character of $\mathrm{Ind}_{H}^{G}\chi_{{\,\mathrm{Id}}}$ is given by

   size:	1	3	3	7	7
   	$e$	$y$	$y^{3}$	$x$	$x^{2}$
   $\mathrm{Ind}_{H}^{G}\chi_{{\,\mathrm{Id}}}$	3	3	3	0	0

   Since all the entries of the above are non-negative, this character has positive inner product with the trivial representation of $G$, and is therefore not irreducible (one can quickly see that it is the direct sum of the three one-dimensional irreducible representations). Next we induce the non-trivial representation of $H$ given by $\sigma_{1}(y^{j})=\eta^{j}$, where $\eta=e^{2\pi i/7}$:

   size:	1	3	3	7	7
   	$e$	$y$	$y^{3}$	$x$	$x^{2}$
   $\mathrm{Ind}_{H}^{G}\sigma_{1}$	3	$\eta+\eta^{2}+\eta^{4}$	$\eta^{3}+\eta^{5}+\eta^{6}$	0	0

   Noting that $\eta^{3}+\eta^{5}+\eta^{6}=\overline{\eta+\eta^{2}+\eta^{4}}$), we have

   $$|\eta+\eta^{2}+\eta^{4}|^{2}=|\eta^{3}+\eta^{5}+\eta^{6}|^{2}=(\eta+\eta^{2}+\eta^{4})(\eta^{3}+\eta^{5}+\eta^{6})=2,$$

   giving $\|\mathrm{Ind}_{H}^{G}\sigma_{1}\|_{G}^{2}=1,$ so this is irreducible. We see that inducing $\sigma_{2}$ (defined through $\sigma_{2}(y)=\eta^{2}$) will give the same representation, whereas inducing $\sigma_{3}$ will swap the two terms $\eta+\eta^{2}+\eta^{4}$ and $\eta^{3}+\eta^{5}+\eta^{6}$:

   size:	1	3	3	7	7
   	$e$	$y$	$y^{3}$	$x$	$x^{2}$
   $\mathrm{Ind}_{H}^{G}\sigma_{3}$	3	$\eta^{3}+\eta^{5}+\eta^{6}$	$\eta+\eta^{2}+\eta^{4}$	0	0

   This completes the table:

   size:	1	3	3	7	7
   	$e$	$y$	$y^{3}$	$x$	$x^{2}$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1	1
   $(\psi_{1},{\mathbb{C}})$	1	1	1	$\omega$	$\omega^{2}$
   $(\psi_{2},{\mathbb{C}})$	1	1	1	$\omega^{2}$	$\omega$
   $\mathrm{Ind}_{H}^{G}\sigma_{1}$	3	$\eta+\eta^{2}+\eta^{4}$	$\eta^{3}+\eta^{5}+\eta^{6}$	0	0
   $\mathrm{Ind}_{H}^{G}\sigma_{3}$	3	$\eta^{3}+\eta^{5}+\eta^{6}$	$\eta+\eta^{2}+\eta^{4}$	0	0

   (One can compute that $z=\eta+\eta^{2}+\eta^{4}=\frac{-1+\sqrt{-7}}{2}$; $z^{2}=\eta^{2}+\eta^{4}+\eta+2\eta^{3}+2\eta^{5}+2\eta^{6}=-2-z$, i.e. $z^{2}+z+2=0$.)