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For $n\ge 1$, the dihedral group $D_n$ is the group generated by two elements $r$, $s$, where:

the element $r$ has order $n$, the element $s$ has order $2$, and the element $sr$ has order $2$.

Its presentation in terms of generators and relations is therefore:

$D_n=\langle r,s\mid r^n=s^2=(sr)^2=1\rangle$, where $1$ is the neutral element.

Which of the following are true?


If $f:D_n\rightarrow G$, $n\ge 3$, is a group homomorphism with $f(sr)=f(1)$, then:

$f(r)$ has order $n/2$.


If $n\ge 3$ is prime and $f:D_n\rightarrow G$ is a group homomorphism with $f(r)\not=f(1)$:

then $f(r)$ has order $n$.


Suppose $d\ge1$ divides $n$ and let $f:D_n\rightarrow D_n$ be the map:

$f(s)=s$, $f(r)=r^{n/d}$

...then we have $f(g)=f(g')$ if and only if we have the equality of cosets:

$g\langle r^d\rangle=g'\langle r^d\rangle$

...where $\langle r^d\rangle$ is the cyclic subgroup of $D_n$ generated by $r^d$.


Let $f:D_n\rightarrow D_n$ be the map:

$f(s)=s$, $f(r)=r^2$



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