HW11

1. The following algebra isomorphisms hold for $R$-algebras $A$ and $B$, where $R$ is a commutative ring.

• (i) $M_n(A){\otimes }_{R}B\simeq M_n\left(A{\otimes }_{R}B\right)$.
• (ii) $M_n(A){\otimes }_R{M}_m(B)\simeq M_{nm}\left(A{\otimes }_{R}B\right)$.

\begin{proof} i) Define

$$\varphi :M_n(A)\times B\to M_n(A{\otimes }_{R}B),((a_{ij}),b)\mapsto (a_{ij}\otimes b).$$

It is easy to check $\varphi$ is $R$-bilinear. By the universal property of tensor products, there exists a unique $R$-module homomorphism $\Phi :M_n(A){\otimes }_{R}B\to M_n(A{\otimes }_{R}B)$ such that $\Phi (M\otimes b)=(a_{ij}\otimes b)$ for any $M=(a_{ij})\in M_n(A)$ and $b\in B$. Since any element in $M_n(A){\otimes }_{R}B$ is a finite sum ${\sum }_k{M}_k\otimes b_k$, we have $\Phi ({\sum }_k{M}_k\otimes b_k)={\sum }_k\Phi (M_k\otimes b_k)$.

We can check that $\Phi$ is a homomorphism between $R$-algebras. Firstly, $\Phi (I_A\otimes 1_B)=I_{A\otimes B}$ is trivial. Furthermore, for any $M_1=(a_{ij}),M_2=(c_{ij})$ and $b_1,b_2\in B$, define $P=M_1{M}_2$, $X=\Phi (M_1\otimes b_1)$ and $Y=\Phi (M_2\otimes b_2)$. Then we have

$$\Phi ((M_1{M}_2)\otimes (b_1{b}_2))=\Phi (P\otimes b_1{b}_2)=(p_{ik}\otimes b_1{b}_2)=\left(\left(\sum _j{a}_{ij}{c}_{jk}\right)\otimes b_1{b}_2\right)$$

and $(XY{)}_{ik}={\sum }_j{x}_{ij}{y}_{jk}={\sum }_j(a_{ij}\otimes b_1)(c_{jk}\otimes b_2)={\sum }_j(a_{ij}{c}_{jk})\otimes b_1{b}_2$. Therefore, $\Phi ((M_1{M}_2)\otimes (b_1{b}_2))=\Phi (M_1\otimes b_1)\Phi (M_2\otimes b_2)$. Now we have proved that $\Phi$ is an $R$-algebra homomorphism.

On the other hand, let $E_{kl}(x)$ be the matrix with $x$ in the $(k,l)$-th position and zeros elsewhere. Define $\Psi (E_{kl}(a\otimes b))=E_{kl}(a)\otimes b$. Since each $X\in M_n(A\otimes B)$ can be written as $X={\sum }_{i,j}{E}_{ij}(x_{ij})$ with $x_{ij}={\sum }_{\text{finite}}{a}_{ijp}\otimes b_{ijp}$, we can define $\Psi$ as

$$\Psi (X)=\Psi \left(\sum _{i,j,p}{E}_{ij}({\alpha }_{ijp}\otimes {\beta }_{ijp})\right)=\sum _{i,j,p}\Psi (E_{ij}({\alpha }_{ijp}\otimes {\beta }_{ijp}))=\sum _{i,j,p}(E_{ij}({\alpha }_{ijp})\otimes {\beta }_{ijp}).$$

It is easy to check that $\Psi$ is an $R$-algebra homomorphism and $\Phi ,\Psi$ are inverses. Now we proved that $M_n(A){\otimes }_{R}B\simeq M_n\left(A{\otimes }_{R}B\right)$.

ii) By i), we have

$$M_n(A){\otimes }_R{M}_m(B)\simeq M_n(A{\otimes }_R{M}_m(B))\simeq M_n(M_m(A{\otimes }_{R}B)),$$

and $M_n(M_m(A{\otimes }_{R}B))\simeq M_{nm}(A{\otimes }_{R}B)$ is trivial. Now we finish the proof. \end{proof}

2. Let ${\mathfrak{S}}_n$ denote the $n$-th symmetric group, the group of all permutations of $\{1,\dots ,n\}$, of order $n!$.

• (i) Check that the set $\{1,(12)(34),(13)(24),(14)(23)\}$ is a normal subgroup of ${\mathfrak{S}}_4$. Deduce that this defines a group morphism ${\mathfrak{S}}_4\to {\mathfrak{S}}_3$.
• (ii) Prove that this morphism is surjective.

\begin{proof} i) It is easy to check the set forms a group. Note that the conjugacy class of ${\mathfrak{S}}_4$ containing $\{(12)(34)\}$ is $\{(12)(34),(13)(24),(14)(23)\}$, then the set is a normal subgroup. Let $K:=\{1,(12)(34),(13)(24),(14)(23)\}$, and let $\pi :{\mathfrak{S}}_4\to {\mathfrak{S}}_4/K$ be a canonical quotient map. We claim that ${\mathfrak{S}}_4/K\simeq {\mathfrak{S}}_3$. Note that the order of ${\mathfrak{S}}_4/K$ is $4!/4=3!$. Recall that the group of order $6$ is isomorphic to $C_6$ or $S_3$, so it suffices to show ${\mathfrak{S}}_4/K$ is non-abelian.

Otherwise, if ${\mathfrak{S}}_4/K$ is abelian, then $K⩾{\mathfrak{S}}_4^{\prime }$, where ${\mathfrak{S}}_4^{\prime }$ is the $4$-th alternating group with order $12$, which is impossible. Therefore, ${\mathfrak{S}}_4/K$ is non-abelian and so there is a group morphism ${\mathfrak{S}}_4\to {\mathfrak{S}}_3$.

ii) By the define of $\pi$ in i), it is a quotient map and so is surjective. \end{proof}

3.

• (i) Let $(X,\rho )$ be a transitive Set-representation of $G$. For $x\in X$, let $G_x$ denote the fixator of $x$ in $G$. Then the map

  $$G/G_x\to X,g{G}_x\mapsto \rho (g)(x)$$

  defines an isomorphism $\left(G/G_x,{\rho }_{G_x}\right)\stackrel{\sim }{\to }(X,\rho )$.

• (ii) For subgroups $H$ and $H^{\prime }$ of $G$, the representations $\left(G/H,{\rho }_H\right)$ and $\left(G/H^{\prime },{\rho }_{H^{\prime }}\right)$ are isomorphic if and only if $H$ and $H^{\prime }$ are conjugate in $G$.

\begin{proof} i) Define

$$\phi :G\to X,g\mapsto \rho (g)x.$$

Note that for any $g_1,g_2\in G$ such that $\phi (g_1)=\phi (g_2)$, there is $g_1{g}_2^{-1}\in G_x$. Hence $\phi$ induces a map

$$\overline{\phi }:G/G_x\to X,g{G}_x\mapsto \rho (g)x,$$

which is injective by the argument above and is surjective by $(X,\rho )$ transitive. Furthermore, $\overline{\phi }$ satisfies $\overline{\phi }(k(g{G}_x))=k\overline{\phi }(g{G}_x)$, so $\overline{\phi }$ is an isomorphism.

(ii) The set-representations $(G/H,{\rho }_H)$ and $(G/H^{\prime },{\rho }_{H^{\prime }})$ are defined by ${\rho }_H(k)(gH)=(kg)H$ and ${\rho }_{H^{\prime }}(k)(g{H}^{\prime })=(kg)H^{\prime }$, respectively.

Suppose $\Phi :G/H\to G/H^{\prime }$ is an isomorphism of $G$-sets. Then for all $k\in G$ and $gH\in G/H$, we have $\Phi (k\cdot (gH))=k\cdot \Phi (gH)$. Assume that $\Phi (H)=g_0{H}^{\prime }$ for some $g_0\in G$, then the stabilizer of $g_0{H}^{\prime }$ is $G_{g_0{H}^{\prime }}=\{k\in G\mid k(g_0{H}^{\prime })=g_0{H}^{\prime }\}$, which is equal to $G_H=H$. Note that

$$G_{g_0{H}^{\prime }}=\{k\in G\mid k(g_0{H}^{\prime })=g_0{H}^{\prime }\}=\{k\in G:g_0^{-1}k{g}_0\in H^{\prime }\}=g_0{H}^{\prime }{g}_0^{-1}.$$

Since the stabilizer of $g_0{H}^{\prime }$ and $H$ are the same, we have $H=g_0{H}^{\prime }{g}_0^{-1}$, that is, $H$ and $H^{\prime }$ are conjugate.

Conversely, assume that $H$ and $H^{\prime }$ are conjugate, then there exists $g_0\in G$ such that $H^{\prime }=g_0^{-1}H{g}_0$. Define $\Phi :G/H\to G/H^{\prime }$ by $\Phi (gH)=g{g}_0{H}^{\prime }$, then it is easy to check $\Phi$ is a well-defined bijection and $k(\Phi (gH))=\Phi (kgH)$. Hence the representations are isomorphic. \end{proof}