# Surjective Homomorphism Dihedral Group

For the part which I have underlined, how do i know these are the only 2 options for groups of size 6?  It follows from the fact that any group of order $2p$ with $p>2$ prime is either cyclic, i.e., isomorphic to $C_{2p}$ or isomorphic to the dihedral group $D_{2p}$. The proof uses Chauchy's theorem, which says that if $p$ divides the order of a group $G$, then there is an element of order $p$ in $G$. For $p=3$ we obtain the two choices $C_6$ and $D_6$.
Proposition: Every group of order $2p$,for an odd prime $p$, is cyclic or dihedral.
Proof: By Cauchy's theorem $G$ contains elements $s$ and $r$ with $ord(s)=2$ and $ord(r)=p$. Let $H=\langle r\rangle$. Then $H$ is of index $2$ in $G$, hence normal. Because $s\not\in H$ we have $G=H\cup Hs=\{ 1,r,\ldots ,r^{p-1},s,rs,\ldots ,r^{p-1}s\}$. Since $H$ is normal, $srs^{-1}=r^j$ for some $j$. Because $s^2=1$ we have $r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{j^2}$. This means $j^2\equiv 1 \mod p$. Because $\mathbb{Z}/p\mathbb{Z}$ is a field, the quadratc equation $j^2=1$ has the solutions $j=\pm 1$. The first case, $j\equiv 1 (p)$ gives a commutative group, $rs=sr$, and hence $G\simeq C_2\times C_p\simeq C_{2p}$, the second case gives $srs^{-1}=r^{-1}$, which is the dihderal group.