Question

Exercise 4.5.3. Let G-(g g 1 be a group of order 2 and V a CG-module of Let u +202 +2,u2 2v1 - 2 +2vs,u vector space spanned by ui, for i-1,2,3 2v - 202 +vs, and hence U the (i) Prove that U is a CG-submodule of V fori 1,2,3, and that (ii) Let λ C and u-ul + U2 + λν3 V. Find the value(s) of λ for which the subspace U spanned by u is a CG-submodule of V, and find all i є {1.2.3} for which UU

Answer 1

i) To show that $U_i$ is $\mathbb CG$ -submodule of , we merely need to show that $gu_i=\lambda_iu_i$ for some $\lambda_i\in\mathbb C$ .

Now, consider $u_1=v_1+2v_2+2v_3$ ; then

$\begin{align*}gu_1&=g(v_1+2v_2+2v_3)\\ &=gv_1+2gv_2+2gv_3\\ &=(-9v_1+4v_2-8v_3)+2(-4v_1+v_2-4v_3)+2(8v_1-4v_2+7v_3)\\ &=-v_1-2v_2-2v_3\\ &=-u_1\end{align*}$

Similarly,

$\begin{align*}gu_2&=g(2v_1-v_2+2v_3)\\ &=2gv_1-gv_2+2gv_3\\ &=2(-9v_1+4v_2-8v_3)-(-4v_1+v_2-4v_3)+2(8v_1-4v_2+7v_3)\\ &=2v_1-v_2+2v_3\\ &=u_2\end{align*}$

and

$\begin{align*}gu_3&=g(2v_1-2v_2+v_3)\\ &=2gv_1-2gv_2+gv_3\\ &=2(-9v_1+4v_2-8v_3)-2(-4v_1+v_2-4v_3)+(8v_1-4v_2+7v_3)\\ &=-2v_1+2v_2-v_3\\ &=-u_3\end{align*}$

Thus, $U_i$ is $\mathbb CG$ -submodule of for $\begin{align*}i=1,2,3\end{align*}$ .

To show that $\begin{align*}V=U_1\oplus U_2\oplus U_3\end{align*}$ it suffices to show that the vectors $\begin{align*}u_1,u_2,u_3\end{align*}$ are linearly independent. Consider the linear transform $\begin{align*}T:V\rightarrow V\end{align*}$ , given by $\begin{align*}T(v_i)=u_i\end{align*}$ for $\begin{align*}i=1,2,3\end{align*}$ . With respect to the basis $\begin{align*}\{v_1,v_2,v_3\}\end{align*}$ its matrix is

$\begin{align*}T=\begin{pmatrix}1&2&2\\ 2&-1&-2\\ 2&2&1 \end{pmatrix}\end{align*}$

Is determinant is

$\begin{align*}\det T&=\begin{pmatrix}-1&-2\\ 2&1 \end{pmatrix}-2\begin{pmatrix}2&-2\\ 2&1 \end{pmatrix}+2\begin{pmatrix}2&-1\\ 2&2\end{pmatrix}\\ &=3-12+12\\ &=3\end{align*}$

Thus, $\begin{align*}T:V\rightarrow V\end{align*}$ is invertible, therefore, it maps the basis $\begin{align*}\{v_1,v_2,v_3\}\end{align*}$ to a basis, which is $\begin{align*}\{u_1,u_2,u_3\}\end{align*}$ by construction.

ii) Again, $\begin{align*}\lambda\end{align*}$ needs to be such that $\begin{align*}gu=\lambda_uu\end{align*}$ for some $\begin{align*}\lambda_u\in\mathbb C\end{align*}$ .

Now,

$\begin{align*}gu&=gv_1+gv_2+\lambda gv_3\\ &=(-9v_1+4v_2-8v_3)+(-4v_1+v_2-4v_3)+\lambda(8v_1-4v_2+7v_3)\\ &=(-9-4+8\lambda)v_1+(4+1-4\lambda)v_2+(-8-4+7\lambda)v_3\\ &=(-13+8\lambda)v_1+(5-4\lambda)v_2+(-12+7\lambda)v_3\end{align*}$

Thus, for $\begin{align*}gu=\lambda_uu\end{align*}$ to hold  for some $\begin{align*}\lambda_u\in\mathbb C\end{align*}$ we need

$\begin{align*}(-13+8\lambda)v_1+(5-4\lambda)v_2+(-12+7\lambda)v_3&=\lambda_uv_1+\lambda_uv_2+\lambda_u\lambda v_3\end{align*}$

This implies, in particular

$\begin{align*}-13+8\lambda=5-4\lambda\\ \Rightarrow\hspace{3cm}12\lambda&=18\\ \Rightarrow\hspace{3.4cm}\lambda&=1.5\end{align*}$

Finally, we check that for this value we do have consistency in

$\begin{align*}(-13+8\lambda)v_1+(5-4\lambda)v_2+(-12+7\lambda)v_3&=\lambda_uv_1+\lambda_uv_2+\lambda_u\lambda v_3\end{align*}$

as the left side is

$\begin{align*}(-13+8\lambda)v_1+(5-4\lambda)v_2+(-12+7\lambda)v_3&=-v_1-v_2-1.5v_3\end{align*}$

and the right side is

$\begin{align*}\lambda_uv_1+\lambda_uv_2+\lambda_u\lambda v_3&=-v_1-v_2-1.5v_3\end{align*}$

Thus, we need $\begin{align*}\lambda=1.5\end{align*}$ .

For the last part, note that all the $\mathbb CG$ -submodules $\begin{align*}U, U_i\end{align*}$ , where $\begin{align*}i=1,2,3\end{align*}$ are one-dimensional (vector) subspaces. Since $\begin{align*}gu=-u\end{align*}$ , and $\begin{align*}gu_i=-u_i\end{align*}$ for $\begin{align*}i=1,3\end{align*}$ , we find that $\begin{align*}U_i \cong U\end{align*}$ for $\begin{align*}i=1,3\end{align*}$ . On the other hand, since $\begin{align*}gu=-u\end{align*}$ , and $\begin{align*}gu_2=u_2\end{align*}$ , we know that $\begin{align*}U_2 \not\cong U\end{align*}$ .