Show that $G= \langle x,y\mid x^2,xyx^{-1}y=e \rangle$ is a semidirect product of two of its subgroups

Question

Show that $G= \langle x,y\mid x^2,xyx^{-1}y=e \rangle$ is a semidirect product of two of its subgroups

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my first attempt was to use the theorem: $G\cong H\rtimes K$ iff $H=$ normal $H\cap K=\{e\}$ and $G=HK$

but I can't show if one of them $(H=\langle y\rangle ,K=\langle x\rangle)$ is normal and also can't show $G=HK$

the hint I get is to take $H=\langle y\rangle ,K=\langle x\rangle$ and $f:K\rightarrow \text{Aut(H)}$

$f(x)=\phi_x$ where $\phi_x(y)=xyx^{-1}=y^{-1}$

I am confused about what I should show with this hint.

Clearly $f $ is a homomorphism, if I take the product $(h,k)(h',k')=(hf(k)h,kk')$ how can I show that $G\cong H\rtimes K$. Can someone explain, this topic is fairly confusing.

Answer 1

The group presentation is the standard presentation of the infinite dihedral group $\langle a,b\mid a^2=1, aba=b^{-1}\rangle$. In that group $H=\langle b\rangle$ is normal and infinite cyclic, $K=\langle a\rangle$ is of order $2$. Thus $H\cap K=\{e\}$ and $HK$ is the whole group, as required.