HW15

1. Prove: Let $G$ be a finite group and $p$ a prime. Each element $x\in G$ can be uniquely written $x=st$ where $s$ is $p$-regular, $t$ is $p$-singular, and $st=ts$. If $x_1=s_1{t}_1$ is such a decomposition of an element $x_1$ that is conjugate to $x$, then $s$ is conjugate to $s_1$, and $t$ is conjugate to $t_1$.

\begin{proof} Assume that $o(x)=n=p^{\ell }\cdot k$ with $(p^{\ell },k)=1$. By Bezout’s theorem, there exist integers $a,b$ such that $a\cdot p^{\ell }+b\cdot k=1$. Then we have $x=x^{a\cdot p^{\ell }+b\cdot k}=x^{a{p}^{\ell }}\cdot x^{bk}$. Define $s=x^{a{p}^{\ell }}$ and $t=x^{bk}$, then $s^k=t^{p^{\ell }}=1$. Thus $(o(s),p)=1$, $o(t)$ is a power of $p$, and $st=ts$. Now we proved that such decomposition exists. For any decomposition $x=st$ satisfying the conditions above, we have $o(x)=o(s)o(t)$ as $(o(s),o(t))=1$. Let $k_s=o(s)$, and let $k_t=o(t)$. Then there exist integer $\alpha ,\beta$ such that $\alpha k_s+\beta k_t=1$. It deduces that $s=s^{\alpha k_s+\beta k_t}=s^{\beta k_t}$ and $t=t^{\alpha k_s+\beta k_t}=t^{\alpha k_s}$. Also note that $x^{\beta k_t}=(st{)}^{\beta k_t}=s^{\beta k_t}=s$ and $x^{\alpha k_s}=(st{)}^{\alpha k_s}=t^{\alpha k_s}=t$. Hence $s,t\in ⟨x⟩$. If such decomposition is not unique, that is, there exists $x=st=s^{\prime }{t}^{\prime }$, then $g:=s^{\prime -1}s=t^{\prime }{t}^{-1}$. Since $s,s^{\prime },t,t^{\prime }\in ⟨x⟩$, they commute and so $o(g)=o(s^{\prime -1}s)=o(t^{\prime }{t}^{-1})$ is both $p$-regular and $p$-singular, which yields that $g=1$, $s^{\prime }=s$, and $t^{\prime }=t$, contradiction.

If $x_1=s_1{t}_1$ is such a decomposition of an element $x_1$ that is conjugate to $x$, then there exists $h\in G$ such that $x=x_1^h$. Then $x=x_1^h=(s_1{)}^h(t_1{)}^h$. Since $s_1{t}_1=t_1{s}_1$, there is $s_1^h{t}_1^h=t_1^h{s}_1^h$. Also $o(s_1)=o(s_1^h)$ and $o(t_1)=o(t_1^h)$. Hence $s_1^h{t}_1^h$ is a decomposition of $x$ and by uniqueness there is $s_1^h=s$, $t_1^h=t$, i.e., $s$ is conjugate to $s_1$, and $t$ is conjugate to $t_1$. \end{proof}

2. Find the complete list of subgroups $H$ of the dihedral group $G=D_8$ of order 8 such that the 2-dimensional simple representation over $\mathbb{C}$ can be written ${\mathrm{Ind}}_H^{G}U$ for some 1-dimensional representation $U$ of $H$.

\begin{proof} Recall that

Let $W$ be a $H$-module with character $\psi$, then $\psi {\uparrow }_H^G(g)=\frac{1}{|H|}{\sum }_{x\in G}\psi (x^{-1}gx)$ with the convention that $\psi (g)=0$ if $g\notin H$.

Link to original

If a $2$-dimensional representation over $\mathbb{C}$ of $D_8$ can be written as ${\mathrm{Ind}}_H^{G}U$ for some 1-dimensional representation $U$ of $H$, then $[G:H]=2$ and so $H⊲G$. Assume that $G=⟨a,b:a^4=b^2=1,a^b=a^{-1}⟩$, then all possibilities of are $H=⟨a⟩\simeq C_4$, $H=⟨a^2⟩\times ⟨b⟩\simeq C_2\times C_2$, and $H=⟨a^2⟩\times ⟨ab⟩\simeq C_2\times C_2$.

The character table of $C_4$ is

	$1$	$a$	$a^2$	$a^3$
${\chi }_1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$i$	$-1$	$-i$
${\chi }_3$	$1$	$-i$	$-1$	$i$
${\chi }_4$	$1$	$-1$	$1$	$-1$

and the character table of $C_2\times C_2$ is

	$1$	$a^2$	$b$	$a^{2}b$
${\psi }_1$	$1$	$1$	$1$	$1$
${\psi }_2$	$1$	$-1$	$1$	$-1$
${\psi }_3$	$1$	$1$	$-1$	$-1$
${\psi }_4$	$1$	$-1$	$-1$	$1$

Notice that

• ${\chi }_2{\uparrow }_{C_4}^G(1)=2$
• ${\chi }_2{\uparrow }_{C_4}^G(a)=\frac{1}{4}{\sum }_{x\in G}\chi (x^{-1}ax)=\chi (a)+\chi (a^{-1})=0$
• ${\chi }_2{\uparrow }_{C_4}^G(a^2)=\frac{1}{4}{\sum }_{x\in G}\chi (x^{-1}{a}^{2}x)=\chi (a^2)+\chi (a^{-2})=-2$
• ${\chi }_2{\uparrow }_{C_4}^G(a^{i}b)=\frac{1}{4}{\sum }_{x\in G}\chi (x^{-1}{a}^{i}bx)=0$ for all $i\in \{0,1,2,3\}$.

Remark that the conjugacy classes of $D_8$ are

$$\{1\},\{a,a^3\},\{a^2\},\{b,a^{2}b\},\{ab,a^{3}b\}.$$

Then one can check $⟨{\chi }_2{\uparrow }_{C_4}^G,{\chi }_2{\uparrow }_{C_4}^G⟩=\frac{1}{8}(2^2\times 1+0^2\times 2+(-2{)}^2\times 1)=1$. Hence ${\chi }_2{\uparrow }_{C_4}^G$ is irreducible. Similarly, we can compute that ${\psi }_2{\uparrow }_{C_2\times C_2}^G$ is irreducible whenever $H=⟨a^2⟩\times ⟨b⟩$ or $H=⟨a^2⟩\times ⟨ab⟩$. Therefore, the complete list of subgroups $H$ is $\{⟨a⟩,⟨a^2⟩\times ⟨b⟩,⟨a^2⟩\times ⟨ab⟩\}$ where $G=D_8=⟨a,b:a^4=b^2=1,a^b=a^{-1}⟩$. \end{proof}

3. Determine all simple $kG$-modules for

• (i) $G=S_3$ and $k$ is the field of 3 elements.
• (ii) $G=A_4$ and $k$ is a field of $2^n$ (with $n\ge 2$) elements.

\begin{proof} i) Since $k$ is the field of $3$ elements, there is $k\simeq {\mathbb{F}}_3$ and so $\mathrm{char}k\mid |G|$.

We claim that ${\mathbb{F}}_3$ is a splitting field for $G$. Recall thats the trivial representation of $O_3(G)\simeq C_3$ is the unique irreducible $k{O}_3(G)$-module, so the number of simple $kG$-module equals the number of simple $kG/O_3(G)$-module. Since $G/O_3(G)\simeq C_2$ and ${\mathbb{F}}_3$ contains $2$-th root of unity, by ^ci9c9b we know $k$ is splitting for $G/O_p(G)$. Now we prove the claim.

Notice that the conjugacy classes of $G/O_p(G)$ are $\{\overline{e}\}$ and $\{\overline{(12)}\}$, so $G/O_p(G)$ has two simple $kG$-modules by ^z05r54. Thus, $G$ also has two simple $kG$-modules.

Consider $1$-dimensional $kG$-modules. For each simple $kG$-module with dimension $1$, it corresponds to a group homomorphism $G\to {\mathbb{F}}_3^{\times }$, whose kernel contains $G^{\prime }$. So it suffices to consider the group homomorphism $G/G^{\prime }\to {\mathbb{F}}_3^{\times }$. Since $G/G^{\prime }=\{\overline{e},\overline{(12)}\}\simeq C_2$, there exist two group homomorphism from $G/G^{\prime }$ to ${\mathbb{F}}_3^{\times }$, and each of them induces a group homomorphism from $G$ to ${\mathbb{F}}_3^{\times }$, that is,

$$\begin{array}{rl}& {\phi }_1:G\to {\mathbb{F}}_3^{\times },g\mapsto 1\text{ for all }g\in G,\\ & {\phi }_2:G\to {\mathbb{F}}_3^{\times },e\mapsto 1,(12)\mapsto 2,(123)\mapsto 1.\end{array}$$

As ${\phi }_1\ne {\phi }_2$ and each of them induces a $1$-dimensional simple $kG$-module (defined by $g\cdot x=\phi (g)x$ for any $g\in G,x\in {\mathbb{F}}_3$), we get all simple ${\mathbb{F}}_{3}G$-modules.

ii) Now we assume that $k={\mathbb{F}}_{2^n}$ with $n⩾2$, so $\mathrm{char}k\mid |G|$. Since $P:=⟨(12)(34),(14)(23)⟩⊲G$ is a Sylow $2$-subgroup, there is $O_2(G)=P$ and so the number of simple $kG$-module equals the number of simple $kG/O_2(G)$-module. Since $G/O_2(G)\simeq C_3$ and $3\mid 2^{2k}-1$ for any $k\in {\mathbb{N}}_{+}$, one can check $k$ is a splitting field for $G/O_2(G)$ when $n$ is even. So we divide the problem into two parts.

Firstly, when $n$ is even, $k$ is a splitting field for $G/O_2(G)$. Then $G/O_2(G)$ has three conjugacy classes and so $G$ has three simple $kG$-modules by ^z05r54. Furthermore, note that there are three group homomorphisms from $G/G^{\prime }$ to ${\mathbb{F}}_{2^n}^{\times }$, that is,

$$\begin{array}{rl}& {\phi }_1:G\to {\mathbb{F}}_{2^n}^{\times },g\mapsto 1\text{ for all }g\in G,\\ & {\phi }_2:G\to {\mathbb{F}}_{2^n}^{\times },e\mapsto 1,(12)(34)\mapsto 1,(123)\mapsto \omega \\ & {\phi }_3:G\to {\mathbb{F}}_{2^n}^{\times },e\mapsto 1,(12)(34)\mapsto 1,(123)\mapsto {\omega }^2.\end{array}$$

Then we can get all simple $kG$-modules by the similar argument in i), and all of them are dimension $1$.

Now we assume that $n$ is odd. It suffices to find simple module of $C_3$ over ${\mathbb{F}}_{2^n}$. As $\mathrm{char}{\mathbb{F}}_{2^n}\nmid |C_3|$, the group algebra ${\mathbb{F}}_{2^n}{C}_3$ is semisimple.

The group algebra ${\mathbb{F}}_{2^n}{C}_3$ is isomorphic to ${\mathbb{F}}_{2^n}[X]/(X^3-1)$. Since $\mathrm{char}{\mathbb{F}}_{2^n}=2$, there is $X^3-1=(X+1)(X^2+X+1)$ and $(X+1,X^2+X+1)=1$. It deduces that

$${\mathbb{F}}_{2^n}{C}_3\simeq {\mathbb{F}}_{2^n}[X]/(X^3-1)\simeq {\mathbb{F}}_{2^n}[X]/(X+1)\oplus {\mathbb{F}}_{2^n}[X]/(X^2+X+1):=S_1\oplus S_2.$$

Note that $S_1={\mathbb{F}}_{2^n}[X]/(X+1)\simeq {\mathbb{F}}_{2^n}$ is a $1$-dimensional module where the generator of $C_3$ acts as $X$ and $X\equiv -1\equiv 1(\mathrm{mod}X+1)$, so it is a trivial module. Similarly, since $X^2+X+1$ is irreducible, $S_2$ is a simple $2$-dimensional module where the generator $c$ of $C_3$ acts as $X$, and the corresponding matrix representation is

$$\rho (c)=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right).$$

Therefore, the $2$-dimensional simple $A_4$-module corresponds the following representation

$$\rho :A_4\to {\mathrm{M}}_2({\mathbb{F}}_{2^n}),e\mapsto \mathrm{id},(12)(34)\mapsto \mathrm{id},(123)\mapsto \left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right).$$

Since ${\mathbb{F}}_{2^n}{C}_3\simeq S_1\oplus S_2$ and ${\mathbb{F}}_{2^n}{C}_3$ is semisimple, there are two simple ${\mathbb{F}}_{2^n}{A}_4$-modules as above. \end{proof}