Prove a CG module is irreducible

[fishturtle1]

Homework Statement

Suppose $G = D_6 = \langle a, b \vert a^3 = b^2 = 1, b^{-1}ab = a^{-1} \rangle$ and let $\omega = e^{2\pi i/3}$. Prove that the $2$ dimensional subspace $W$ of the $\mathbb{C}G$ defined by
$$W = sp(1 + \omega^2a + \omega a^2, b + \omega^2ab + \omega a^2b)$$

is an irreducible $\mathbb{C}G$-submodule of the regular $\mathbb{C}G$ module.

Relevant Equations

Definitions:

We define $\mathbb{C}G = \lbrace \sum_{g \in G} \mu_g g : \mu_g \in \mathbb{C} \rbrace$. For $r \in \mathbb{C}G$ and $h \in G$, we define $rh = \sum_{g \in G} \mu_g (gh)$.

We say $W$ is a $\mathbb{C}G$ submodule of $\mathbb{C}G$ if it is a subspace of $\mathbb{C}G$ and is satisfies $wh \in W$ for all $w \in W$ and $h \in G$.

We say a $\mathbb{C}G$ module is irreducible if its only submodules are itself and $\lbrace 0 \rbrace$.

Proof: We first show $W$ is a $\mathbb{C}G$ module. Let $u = 1 + \omega^2a + \omega a^2$ and $v = b + \omega^2ab + \omega a^2b$. It is enough to show $ua, ub, va, vb \in W$. We have\begin{align}
ua &= a + \omega^2a^2 + \omega \\
ub &= b + \omega^2ab + \omega a^2b \\
va &= ba + \omega^2 aba + \omega a^2ba = a^3b + \omega^2 b + \omega^2 ab\\
vb &= 1 + \omega^2a + \omega a^2 \\
\end{align}

Then $ub = v \in W$ and $vb = u \in W$. But I do not see how to show $ua$ and $va$ are in $W$. If I could do this, then I would have to show $W$ has no non trivial submodule. Suppose $U$ is a submodule of $W$ such that $\dim U = 1$. Then $U = sp( \alpha u + \beta v)$ for some $\alpha, \beta \in \mathbb{C}$. Since $U$ is a $\mathbb{C}G$ module, we must have $xg \in U$ for all $x \in U$ and $g \in G$. In particular, we have $(\alpha u + \beta v )b = \alpha v + \beta u$. So, there exists $\lambda \in \mathbb{C}$ such that $\lambda(\alpha v + \beta u) = \alpha u + \beta v$. We can rewrite this as

$$(\lambda\beta)u + (\lambda\alpha)v = \alpha u + \beta v$$

Since $u, v$ are linearly independent, we have $\lambda \beta = \alpha$ and $\lambda \alpha = \beta$. And this implies $\lambda\beta = \lambda^{-1}\beta$. So, $\lambda = \lambda^{-1}$ or $\beta = 0$. Does this seem on the right track?

 

[fresh_42]

Homework Statement:: Suppose $G = D_6 = \langle a, b \vert a^3 = b^2 = 1, b^{-1}ab = a^{-1} \rangle$ and let $\omega = e^{2\pi i/3}$. Prove that the $2$ dimensional subspace $W$ of the $\mathbb{C}G$ defined by
$$W = sp(1 + \omega^2a + \omega a^2, b + \omega^2ab + \omega a^2b)$$
is an irreducible $\mathbb{C}G$-submodule of the regular $\mathbb{C}G$ module.
Relevant Equations:: Definitions:

We define $\mathbb{C}G = \lbrace \sum_{g \in G} \mu_g g : \mu_g \in \mathbb{C} \rbrace$. For $r \in \mathbb{C}G$ and $h \in G$, we define $rh = \sum_{g \in G} \mu_g (gh)$.

We say $W$ is a $\mathbb{C}G$ submodule of $\mathbb{C}G$ if it is a subspace of $\mathbb{C}G$ and is satisfies $wh \in W$ for all $w \in W$ and $h \in G$.

We say a $\mathbb{C}G$ module is irreducible if its only submodules are itself and $\lbrace 0 \rbrace$.

Proof: We first show $W$ is a $\mathbb{C}G$ module. Let $u = 1 + \omega^2a + \omega a^2$ and $v = b + \omega^2ab + \omega a^2b$. It is enough to show $ua, ub, va, vb \in W$. We have
\begin{align}
ua &= a + \omega^2a^2 + \omega \\
ub &= b + \omega^2ab + \omega a^2b \\
va &= ba + \omega^2 aba + \omega a^2ba = a^3b + \omega^2 b + \omega^2 ab\\
vb &= 1 + \omega^2a + \omega a^2 \\
\end{align}

Then $ub = v \in W$ and $vb = u \in W$. But I do not see how to show $ua$ and $va$ are in $W$.

\begin{align*}
ua&=a+\omega^2a^2+\omega=\omega u\\
va&=uba=ua^{-1}b^{-1}=ua^2b=\omega uab=\omega^2ub=\omega^2v\\
\end{align*}

If I could do this, then I would have to show $W$ has no non trivial submodule. Suppose $U$ is a submodule of $W$ such that $\dim U = 1$. Then $U = sp( \alpha u + \beta v)$ for some $\alpha, \beta \in \mathbb{C}$. Since $U$ is a $\mathbb{C}G$ module, we must have $xg \in U$ for all $x \in U$ and $g \in G$. In particular, we have $(\alpha u + \beta v )b = \alpha v + \beta u$. So, there exists $\lambda \in \mathbb{C}$ such that $\lambda(\alpha v + \beta u) = \alpha u + \beta v$. We can rewrite this as

$$(\lambda\beta)u + (\lambda\alpha)v = \alpha u + \beta v$$

Since $u, v$ are linearly independent, we have $\lambda \beta = \alpha$ and $\lambda \alpha = \beta$. And this implies $\lambda\beta = \lambda^{-1}\beta$. So, $\lambda = \lambda^{-1}$ or $\beta = 0$. Does this seem on the right track?

Yes. This means you have $\beta =0$ or $\lambda =\pm 1$. Now you need to find an argument why the latter cannot be, and do the same with $\alpha .$

 

[fishturtle1]

\begin{align*}
ua&=a+\omega^2a^2+\omega=\omega u\\
va&=uba=ua^{-1}b^{-1}=ua^2b=\omega uab=\omega^2ub=\omega^2v\\
\end{align*}
Yes. This means you have $\beta =0$ or $\lambda =\pm 1$. Now you need to find an argument why the latter cannot be, and do the same with $\alpha .$

Thank you!

Continuing from above, we have $\beta = 0$ or $\lambda = \pm 1$. We will show the latter is not possible. If $\lambda = 1$, then $\beta u + \alpha v = \alpha u + \beta v$. This implies $(\beta - \alpha)u + (\alpha - \beta)v = 0$. Since $u, v$ are linearly independent, we have $\alpha = \beta$. This implies $U = sp(u + v)$. But this implies $(u+v)a = ua + va = \omega u + \omega^2 v \in U$. Since $\omega u + \omega^2 v$ is not a scalar multiple of $U$, we have reached a contradiction.

If $\lambda = -1$, then $-\beta u + -\alpha v = \alpha u + \beta v$. This implies $(-\alpha - \beta)u + (-\alpha - \beta)v = 0$. Since $u, v$ are linearly independent, we have $\alpha = -\beta$. This implies $U = sp(-u + v)$. But this implies $(-u+v)a = -ua + va = -\omega u + \omega^2v \in U$. Since $-\omega u + \omega^2v$ is not a scalar multiple of $-u+v$, we have reached a contradiction.

We can conclude $\beta = 0$. By a similar argument, we can show $\alpha = 0$. We can conclude $U = \lbrace 0 \rbrace$, which contradicts our assumption that $\dim U = 1$. This shows $W$ is irreducible.

 

[fresh_42]

Thank you!

Continuing from above, we have $\beta = 0$ or $\lambda = \pm 1$. We will show the latter is not possible. If $\lambda = 1$, then $\beta u + \alpha v = \alpha u + \beta v$. This implies $(\beta - \alpha)u + (\alpha - \beta)v = 0$. Since $u, v$ are linearly independent, we have $\alpha = \beta$. This implies $U = sp(u + v)$. But this implies $(u+v)a = ua + va = \omega u + \omega^2 v \in U$. Since $\omega u + \omega^2 v$ is not a scalar multiple of $U$, we have reached a contradiction.

If $\lambda = -1$, then $-\beta u + -\alpha v = \alpha u + \beta v$. This implies $(-\alpha - \beta)u + (-\alpha - \beta)v = 0$. Since $u, v$ are linearly independent, we have $\alpha = -\beta$. This implies $U = sp(-u + v)$. But this implies $(-u+v)a = -ua + va = -\omega u + \omega^2v \in U$. Since $-\omega u + \omega^2v$ is not a scalar multiple of $-u+v$, we have reached a contradiction.

We can conclude $\beta = 0$. By a similar argument, we can show $\alpha = 0$. We can conclude $U = \lbrace 0 \rbrace$, which contradicts our assumption that $\dim U = 1$. This shows $W$ is irreducible.

Looks fine. Just one final question: Why are $u,v$ linear independent?

 

[fishturtle1]

Looks fine. Just one final question: Why are $u,v$ linear independent?

I think that is because $W = sp(u, v)$ and we are given $\dim W = 2$. But i think that is circular?

Suppose there are scalars $\lambda_1, \lambda_2$ such that $\lambda_1 u + \lambda_2 v = 0$ where $\lambda_1 \neq 0$ or $\lambda_2 \neq 0$. Multiplying by $b$ on both sides gives $\lambda_1 v + \lambda_2 u = 0$. WLOG, suppose $\lambda_1 \neq 0$. Then $u + \lambda_1^{-1}\lambda_2v = 0$ and $v + \lambda_1^{-1}\lambda_2u = 0$. This implies $u$ and $v$ are scalar multiples of each other.

But any scalar multiple of $u$ looks like $\lambda u = \lambda + \lambda\omega^2a + \lambda \omega a^2$ i.e. it has a constant term. And $v$ does not have a constant term. We can conclude $u$ and $v$ are not scalar multiples of each other, which implies $\lambda_1 = 0 = \lambda_2$?

I am not sure.

 

[fresh_42]

But any scalar multiple of $u$ looks like $\lambda u = \lambda + \lambda\omega^2a + \lambda \omega a^2$ i.e. it has a constant term. And $v$ does not have a constant term.

Yes. That is the argument.

We obviously have $u,v\neq 0$ so $0=\mu u+\nu v$ can be written as $u=\lambda v$ (or the other way around, or $\mu=\nu=0$ in which case we are done). $\lambda \in \mathbb{C}$ so comparing the scalar component yields $\lambda \cdot 1 =0$ which implies $u=0.$ But $u\neq 0.$