# Show that $\rho$ must be 2-dimensional

Let $G=D_8=\langle g,h |g^4=h^2=1, hgh=g^{-1} \rangle$.

One can show that $G$ has $4$ $1$-dimensional representations.

From first principles (no character theory). Suppose $\rho$ is an irreducible representation of $G$ of dimension $>1$. By considering the effect of $\rho(h)$ on an eigenvector $\rho(g)$, show that $\rho$ must be $2$-dimensional.

I have tried:

We know that $\rho(g)v=\lambda v$.

So I composed this with $\rho(h)$, to get

$$\rho(h)\rho(g)v=\rho(h) \lambda v= \lambda \rho(h) v$$

$$\rho(hg)v=\lambda \rho(h) v$$

$$\rho(hg^{-1})v=\lambda \rho(h) v$$

but this is not going anywhere it seems.

HINTS ONLY.

• Try looking at $\rho(g)\rho(h)v$ instead. – მამუკა ჯიბლაძე Apr 11 '15 at 12:00
• How would that help? – Permian Apr 12 '15 at 10:03
• Well I hope it is still only a hint but it would show that $\rho(h)v$ is also a $\rho(g)$-eigenvector – მამუკა ჯიბლაძე Apr 12 '15 at 10:07
• The space is spanned by two vectors $v$ and $\rho(h)v$; if they are linearly dependent, it is in fact 1-dimensional; if not, then it is 2-dimensional. – მამუკა ჯიბლაძე Apr 12 '15 at 10:22
• (This was strictly speaking the argument that any (irreducible or not) nontrivial representation contains a nontrivial invariant subspace of dimension $\leqslant2$) – მამუკა ჯიბლაძე Apr 12 '15 at 10:26

Actually any nontrivial representation of $D_{2n}:=\left\langle g,h\mid g^n=h^2=1, hgh=g^{-1}\right\rangle$ contains a nontrivial subrepresentation of dimension $\leqslant2$.

Any nontrivial representation contains a nonzero $\rho(g)$-eigenvector $v$. Then it is easy to see that $\bar v:=\rho(h)v$ is also a $\rho(g)$-eigenvector. Thus the $\leqslant\!2$-dimensional subspace $\left\langle v,\bar v\right\rangle$ is $\rho(g)$-invariant. But it is obviously also $\rho(h)$-invariant, hence a subrepresentation.

(This is essentially the same as the earlier answer, I just tried to make it as concise as possible)

We have $g^h=g^{-1}$. If $\lambda$ is an eigen value of $p(g)$ it is also eigen value of $p(g^{-1})=p(g^h)$ as they are similar. Thus, we have $\lambda=\dfrac{1}{\lambda}\implies \lambda=+,-1$.

That means that $p(g)v=v$ or $p(g)v=-v$. Hence, the supspace $<v>$ is invariant the under $H=<g>$. Now, show that $W=<v,p(h)v>$ is invariant under $G=D_8$.

• @sandstone: I updated now. – mesel Apr 11 '15 at 12:16
• I cant see where this is going to go – Permian Apr 11 '15 at 19:32
• @sandstone: Can you show that $W$ is $G$ invariant ? – mesel Apr 11 '15 at 19:46