1

Suppose that $f: D_{18} \to GL(2,\Bbb R)$ is a homomorphism, $\lvert r \rvert = 18$ and $f(r) = R := \begin{pmatrix}1& 1 \\ -1& 0\end{pmatrix}$. What is $\lvert \ker(f) \rvert$?

Attempt: I previously showed that if $s$ is the reflection $$f(s) = S := \begin{pmatrix}a &b \\ b-a &-a\end{pmatrix}$$ where $a^2 -ab + b^2 = 1$.

$D_{18} = \{e,s, sr, ..., sr^{17}, r, ..., r^{17}\}$.

Since $f$ is a homomorphism $f(e)=e$ so $e \in \ker(f)$. $f(sr^n) = f(s)f(r)^n = SR^n$ so we can compute whether or not this gives the identity for each $SR^n$ and $R^n$ But is there a more general way than this tedious method?

Doodger24
  • 23
Jeffrey
  • 537

1 Answers1

1

We know that $ker(f)$ is normal subgroup. The possible normal subgroups of $D_{18}$ are $<e>,<r>,<r^2>,<r^3>,<r^6>,<r^9>,<r^2,s>,<r^2,rs>$ (see). Since $f(r^6)=I$(Identity matrix). Therefore $ker f=\{r^6,r^{12},e\}$

Community
  • 1
user177523
  • 745