I am doing the following problem:
Let $G$ be a group such that $|G|=2p$ with $p \geq 3$ prime, then $G$ is abelian or $G \cong D_{2p}$.
Suppose $G$ is not abelian. By Cauchy theorem, there exist $x,y \in G$ with $ord(x)=2$ and $ord(y)=p$. Now, since $|\langle y \rangle |=p$, we have $[G:\langle y \rangle]=\dfrac{|G|}{|\langle y \rangle|}=2$ so $\langle y \rangle \lhd G$. It follows $\langle y \rangle \langle x \rangle =G$.
If I could show $yx=xy^{-1}$, then I could easily show that there exists an isomorphism between $G$ and $D_{2p}$. Since $\langle y \rangle \lhd G$, $xyxy \in \langle y \rangle$. This means $xyxy=y^j$ for some $0\leq j \leq p-1$. I would like to show $j=0$ but at this point I got stuck. I would appreciate help with this part. Thanks in advance.
As shown $\langle y\rangle\trianglelefteq G$, and $\langle y\rangle \langle x\rangle=G$. This can be combined as follows: $G$ is generated by $x,y$, and $\langle y\rangle$ is normalized by both $y$ and $x$. Normalization by $x$ says: $x^{-1}\langle y\rangle x=\langle y\rangle$; so $x^{-1}yx=y^i$, $i\neq 1$ (mod $p$). Then $x^{-2}yx^2=x^{-1}y^ix=y^{i^2}$, i.e. $1^{-1}.y.1=y^{i^2}$, so $i^2=1$ mod $p$. As $p\neq 2$, we have $i=1,-1$ mod $p$, (i.e. $i=1$ or $i=p-1$). As $G$ is non-abelian, $i$ must be $-1$, so $G$ becomes Dihedral.